Dual Nature Of Matter And Radiation – JEE Main PhysicsPractice Questions & PYQs

When a particle with charge $q$ and mass $m$ accelerates through a potential difference $V$, it acquires kinetic energy equal to $qV$, which relates to its momentum $p$ as $qV = \frac{p^2}{2m}$. Rearranging this reveals that momentum is expressed as $p = \sqrt{2mqV}$. Inserting this into the de Broglie relation $\lambda = \frac{h}{p}$ gives the wavelength formula $\lambda = \frac{h}{\sqrt{2mqV}}$. Since the electron and proton possess the same charge magnitude and experience the same potential difference, all variables except mass are identical for both, leading to the inverse relationship $\lambda \propto \frac{1}{\sqrt{m}}$. Consequently, the ratio of their wavelengths is $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$.

The de Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$, which shows that the wavelength is inversely proportional to the square root of the potential. An increase in the wavelength by 50% means the new wavelength is $\lambda_2 = 1.5 \lambda_1$, yielding the ratio $\frac{\lambda_2}{\lambda_1} = \frac{3}{2}$. Equating this to the ratio of the square roots of the potentials gives $\sqrt{\frac{V_1}{V_2}} = \frac{3}{2}$, and squaring both sides results in the ratio $\frac{V_1}{V_2} = \frac{9}{4}$. By comparing this result to the expression $\frac{9}{\alpha}$, it is clear that the value of $\alpha$ is 4.

According to Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is given by $eV_s = \frac{hc}{\lambda} - \phi$, where $V_s$ is the stopping potential, $\lambda$ is the wavelength of the incident light, and $\phi$ is the work function of the metal. For the two cases provided, the equations are $eV_{s1} = \frac{hc}{\lambda} - \phi$ and $eV_{s2} = \frac{hc}{2\lambda} - \phi$. Subtracting the second equation from the first allows the work function to be eliminated, resulting in $e(V_{s1} - V_{s2}) = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$.

Solving for the wavelength yields $\lambda = \frac{hc}{2e(V_{s1} - V_{s2})}$. Substituting the given values leads to $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2 \times 1.6 \times 10^{-19} \times (3.2 - 0.7)} = \frac{19.89 \times 10^{-26}}{2 \times 1.6 \times 10^{-19} \times 2.5} = \frac{19.89 \times 10^{-26}}{8 \times 10^{-19}}$, which simplifies to $\lambda \approx 2.486 \times 10^{-7}$ m, consistent with $2.5 \times 10^{-7}$ m.

According to Einstein's photoelectric equation, the maximum kinetic energy is defined as $K_{max} = h\nu - \Phi$, where the highest incident frequency determines the kinetic energy of the ejected photoelectrons. The wave equation indicates two angular frequencies, $\omega_1 = 3 \times 10^{15} \, \text{rad/s}$ and $\omega_2 = 12 \times 10^{15} \, \text{rad/s}$, so the calculation requires the use of the higher frequency $\omega_2$ to find the maximum energy.

The photon energy in electron volts is determined by converting the angular frequency to ordinary frequency and dividing by the elementary charge
$E = \frac{h \omega_2}{2\pi e} = \frac{6.6 \times 10^{-34} \times 12 \times 10^{15}}{2 \times 3.14 \times 1.6 \times 10^{-19}} \approx 7.88 \, \text{eV}$
Applying Einstein's photoelectric equation, the maximum kinetic energy is obtained by subtracting the work function from this photon energy
$K_{max} = 7.88 \, \text{eV} - 2.8 \, \text{eV} = 5.08 \, \text{eV}$
Rounding this result gives approximately $5.1 \, \text{eV}$.

The total power $P$ emitted by a laser is equal to the number of photons emitted per second $n$ multiplied by the energy of a single photon $E$, represented as $P = n \times E$. Substituting the energy formula $E = \frac{hc}{\lambda}$ into the power equation yields $P = n \times \frac{hc}{\lambda}$, which rearranges to $n = \frac{P \lambda}{hc}$.

Using the given values $P = 6 \times 10^{-3} \text{ W}$, $\lambda = 663 \times 10^{-9} \text{ m}$, $h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$, and $c = 3 \times 10^8 \text{ m/s}$:

$n = \frac{6 \times 10^{-3} \times 663 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8}$

$n = \frac{6 \times 10^{-3} \times 100 \times 6.63 \times 10^{-9}}{6.63 \times 3 \times 10^{-26}}$

$n = \frac{600 \times 10^{-12}}{3 \times 10^{-26}}$

$n = 2 \times 10^{16}$

The de Broglie wavelength $\lambda$ for a particle of mass $m$ and kinetic energy $K$ is defined by $\lambda = \frac{h}{\sqrt{2mK}}$. Using a deuteron mass $m_d = 2m_p$ with kinetic energy $E$ and an alpha particle mass $m_\alpha = 4m_p$ with kinetic energy $2E$, the ratio of their wavelengths is:
$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha K_\alpha}{m_d K_d}} = \sqrt{\frac{4m_p \times 2E}{2m_p \times E}}$
$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{8}{2}} = 2$
The value of $n$ is 2.

Einstein's photoelectric equation governs the interaction, stating that the energy of an incident photon is partitioned between the metal's work function and the maximum kinetic energy of the emitted photo-electrons. This relationship is expressed as $\frac{hc}{\lambda} = \phi + e V_s$, where the stopping potential energy accounts for the kinetic energy of the electrons. Rearranging this equation to determine the work function gives $\phi = \frac{hc}{\lambda} - e V_s$.

The energy of the incoming photon is calculated by substituting the provided constants, which results in $\frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}$, simplifying to $6 \times 10^{-19}$ Joules. The kinetic energy component from the stopping potential is $e V_s = 1.6 \times 10^{-19} \times 0.2$, which equals $0.32 \times 10^{-19}$ Joules. Subtracting this kinetic energy from the photon energy yields the work function $\phi = (6 - 0.32) \times 10^{-19}$ Joules, or $5.68 \times 10^{-19}$ Joules. Given the form $\phi = \alpha \times 10^{-19}$ Joules, it follows that $\alpha = 5.68$.

The De Broglie wavelength of an electron is inversely proportional to its velocity as described by the equation $\lambda = \frac{h}{mv}$. Because the electron's velocity is reduced by $20\%$ in the medium, the new velocity becomes $0.8v$. Substituting this value into the expression yields $\lambda' = \frac{h}{m(0.8v)} = \frac{1}{0.8} \left( \frac{h}{mv} \right)$, which simplifies to $1.25$ times the initial wavelength.

Einstein's photoelectric equation dictates that the maximum kinetic energy of a photoelectron is the difference between the energy of the incident photon and the material's work function, represented as $K = \frac{hc}{\lambda} - \phi$. Because photon energy is inversely proportional to wavelength, the light with the shorter wavelength, $\lambda_2$, provides more energy than the light with the longer wavelength, $\lambda_1$.

Starting with the two given conditions, we have $K_1 = \frac{hc}{\lambda_1} - \phi$ and $K_2 = \frac{hc}{\lambda_2} - \phi$. Using the provided relationship $\lambda_1 = 2\lambda_2$, the energy of the first photon becomes $\frac{hc}{\lambda_1} = \frac{hc}{2\lambda_2}$, which is exactly half of the energy of the second photon. We can rewrite the second condition as $\frac{hc}{\lambda_2} = K_2 + \phi$ and substitute this into the expression for $K_1$, resulting in $K_1 = \frac{1}{2}(K_2 + \phi) - \phi$. Multiplying by two removes the fraction to give $2K_1 = K_2 + \phi - 2\phi$, which simplifies to $2K_1 = K_2 - \phi$. Rearranging this final equation isolates the work function as $\phi = K_2 - 2K_1$.

The power output of an electromagnetic source is equivalent to the total energy emitted per second, which is calculated as the product of the number of photons emitted per second and the energy of an individual photon. Given that a photon's energy is expressed as $E = \frac{hc}{\lambda}$, the relationship between the total power $P$, the number of photons per second $n$, and the wavelength $\lambda$ is $P = n \left( \frac{hc}{\lambda} \right)$.

Rearranging this formula leads to $\lambda = \frac{n \cdot h \cdot c}{P}$. Substituting the provided values of $n = 2.5 \times 10^{22} \text{ photons/s}$, $h = 6.6 \times 10^{-34} \text{ J}\cdot\text{s}$, $c = 3 \times 10^8 \text{ m/s}$, and $P = 15 \times 10^3 \text{ W}$ yields the equation:

$\lambda = \frac{2.5 \times 10^{22} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{15 \times 10^3}$

The calculation produces a wavelength of $\lambda = 3.3 \times 10^{-7} \text{ m}$, or $330 \text{ nm}$. Since this wavelength falls below the visible spectrum, which ranges from approximately $400 \text{ nm}$ to $700 \text{ nm}$, the emitted radiation belongs to the ultraviolet region.

The de Broglie wavelength for a gas molecule is determined by the relation $\lambda = \frac{h}{\sqrt{3mk_B T}}$, where $T$ is the absolute temperature in Kelvin. Given that $T = 27 + 273 = 300$ K, the values are substituted into the equation as follows:

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}}$

Evaluating the term inside the square root yields:

$\sqrt{3 \times 5.31 \times 1.38 \times 300 \times 10^{-49}} = \sqrt{6.595 \times 10^{-46}} = 2.568 \times 10^{-23}$

Dividing the numerator by the result gives:

$\lambda = \frac{6.63 \times 10^{-34}}{2.568 \times 10^{-23}} \approx 2.58 \times 10^{-11} \text{ m}$

Expressing this result in the required format results in $\lambda = 25.8 \times 10^{-12}$ m, which corresponds to the value 26.

The maximum kinetic energy of emitted photoelectrons is determined by Einstein's photoelectric equation, expressed as $KE_{max} = h\nu - h\nu_{th}$, where $h\nu$ is the energy of the incident photon and $\nu_{th}$ is the threshold frequency of the metal. For a fixed incident radiation, the kinetic energy is maximized when the threshold frequency, which represents the energy required to liberate an electron, is minimized. Since $X_1$ corresponds to the lowest threshold frequency on the provided graph, it leaves the greatest portion of the incident photon's energy available as kinetic energy for the photoelectrons.

Einstein's photoelectric equation relates the maximum kinetic energy of photoelectrons to the energy of incident photons and the work function as $K_{\text{max}} = \frac{h\omega}{2\pi} - \Phi$, where $\omega$ is the angular frequency. Since the kinetic energy is zero, the incident energy is exactly equal to the work function, leading to the condition $\frac{h\omega}{2\pi} = \Phi$, or $\omega = \frac{2\pi\Phi}{h}$. Substituting the provided values yields the following calculation:

$\omega = \frac{2 \times 3.14 \times 110 \times 10^{-20}}{6.63 \times 10^{-34}} \approx 1.04 \times 10^{16}$

The angular frequency is 1.04 × 10^{16} rad/s.

The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the expression $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the mass, and $q$ is the charge of the particle. Substituting the provided values $h = 6.6 \times 10^{-34}$ J $\cdot$ s, $m = 6 \times 10^{-27}$ kg, $q = 3 \times 10^{-19}$ C, and $V = 1.21$ V into the equation results in $\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 6 \times 10^{-27} \times 3 \times 10^{-19} \times 1.21}}$ meters. Calculating the denominator gives $\sqrt{36 \times 1.21 \times 10^{-46}}$, which simplifies to $\sqrt{43.56} \times 10^{-23}$ or $6.6 \times 10^{-23}$. Dividing the numerator by the denominator yields $\lambda = 1 \times 10^{-11}$ meters. Expressing this result in the format $\alpha \times 10^{-12}$ meters gives $10 \times 10^{-12}$ meters, which identifies the value of $\alpha$ as 10.

Einstein's photoelectric equation states that the energy of an incident photon is equal to the sum of the metal's work function $\phi$ and the maximum kinetic energy of the photoelectrons, which is expressed as $eV_s$ given a stopping potential $V_s$. In the first scenario with incident light of wavelength $\lambda$ and stopping potential $3V_0$, the relationship is $\frac{hc}{\lambda} - \phi = 3eV_0$, while in the second scenario with wavelength $2\lambda$ and stopping potential $V_0$, it becomes $\frac{hc}{2\lambda} - \phi = eV_0$. By substituting the expression for $eV_0$ from the second equation into the first, we obtain $\frac{hc}{\lambda} - \phi = 3 \left( \frac{hc}{2\lambda} - \phi \right)$. Expanding this expression yields $\frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi$, which simplifies to $2\phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$, resulting in a work function of $\phi = \frac{hc}{4\lambda}$. Since the threshold wavelength $\lambda_0$ is defined by the relation $\phi = \frac{hc}{\lambda_0}$, comparing these expressions confirms that $\lambda_0 = 4\lambda$, identifying the value of $\alpha$ as 4.

Einstein's photoelectric equation serves as the foundation here, balancing the energy of an incoming photon with the energy required to dislodge an electron from the metal surface and the resulting kinetic energy of that electron. The maximum kinetic energy is equivalent to the stopping potential when expressed in electron-volts, which allows us to construct the energy equation as $\frac{hc}{\lambda} = \phi + e V_S$.

Substituting the given values into this relationship, the total energy of the incident electromagnetic wave is the sum of the work function and the stopping potential, calculated as $2.14 \text{ eV} + 2 \text{ eV} = 4.14 \text{ eV}$. Using the provided constant $hc = 1242 \text{ eVnm}$, we determine the wavelength by solving $\lambda = \frac{1242}{4.14}$, which results in 300 nm.

The power emitted by a light source represents the total energy released per unit time, calculated as the product of the number of photons emitted per second and the energy of a single photon, given by $P = N \frac{hc}{\lambda}$. Since the ratio of the power from source $S_1$ to that of source $S_2$ is $2$, we equate the expression to $\frac{N_1 hc / \lambda_1}{N_2 hc / \lambda_2} = 2$. Canceling the constants $h$ and $c$ reduces this to $\frac{N_1 \lambda_2}{N_2 \lambda_1} = 2$. Substituting the values $N_1 = 2 \times 10^{15}$, $\lambda_1 = 600$ nm, and $\lambda_2 = 300$ nm gives $\frac{2 \times 10^{15} \times 300}{N_2 \times 600} = 2$. Simplifying $\frac{300}{600}$ to $\frac{1}{2}$ yields $\frac{2 \times 10^{15}}{2 N_2} = 2$, which further simplifies to $\frac{10^{15}}{N_2} = 2$. Isolating $N_2$ results in $N_2 = 5 \times 10^{14}$ photons per second.

The de Broglie hypothesis links the wavelength $\lambda$ of a particle to its kinetic energy $K$ through the relation $\lambda = \frac{h}{\sqrt{2mK}}$, where $h$ is Planck's constant and $m$ is the particle's mass. By squaring both sides of this equation, we obtain $\lambda^2 = \frac{h^2}{2mK}$, which rearranges to the form $\frac{1}{K} = \left( \frac{2m}{h^2} \right) \lambda^2$. Because the term in parentheses remains constant for a given particle, a graph plotting $\frac{1}{K}$ against $\lambda^2$ generates a straight line passing through the origin, which matches the behavior shown in D.

The photoelectric effect defines the stopping potential ($V_s$) as the specific reverse voltage required to halt the most energetic photoelectrons ejected from a metal surface. Einstein's photoelectric equation relates these quantities through the expression $(K)_{\text{max}} = eV_s$, where $(K)_{\text{max}}$ is the maximum kinetic energy of the emitted electrons and $e$ is the elementary charge. By isolating $V_s$, it becomes clear that $V_s = \frac{(K)_{\text{max}}}{e}$, confirming that the stopping potential is $\frac{1}{e}$ times the maximum kinetic energy of the emitted photoelectrons.

The relationship between the energy and wavelength of a photon is defined by the expression $E = \frac{hc}{\lambda}$, which means its wavelength is represented as $\lambda = \frac{hc}{E}$. For a non-relativistic proton, the kinetic energy relates to its momentum $p$ through the equation $E = \frac{p^2}{2m_p}$, allowing us to express the momentum as $p = \sqrt{2m_p E}$. By applying the de Broglie relation $\lambda_p = \frac{h}{p}$, the wavelength of the proton becomes:
$\lambda_p = \frac{h}{\sqrt{2m_p E}}$

To determine the ratio of the proton's de Broglie wavelength to the wavelength of the photon, we divide the expression for $\lambda_p$ by the expression for $\lambda$:
$\frac{\lambda_p}{\lambda} = \frac{h}{\sqrt{2m_p E}} \times \frac{E}{hc}$

Canceling the Planck constant $h$ from the numerator and denominator simplifies the expression to $\frac{E}{c\sqrt{2m_p E}}$. By incorporating $E$ into the square root, we arrive at the final ratio:
$\frac{\lambda_p}{\lambda} = \frac{1}{c} \sqrt{\frac{E}{2m_p}}$