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

The wave nature of a particle is described by the de Broglie wavelength, which is inversely related to its momentum. Because momentum is linked to kinetic energy through the relationship $p = \sqrt{2m(K.E.)}$, the wavelength is defined as:

$\lambda = \frac{h}{\sqrt{2m(K.E.)}}$

This expression demonstrates that the wavelength is inversely proportional to the square root of the kinetic energy, written as $\lambda \propto \frac{1}{\sqrt{K.E.}}$. Consequently, when the kinetic energy is doubled, the new wavelength scales by a factor of $\frac{1}{\sqrt{2}}$.

The photoelectric effect occurs when incident photons possess energy equal to or greater than the work function of a material. To determine the longest possible wavelength for this emission, we consider the threshold condition where the photon energy exactly matches the work function ($\phi$), which is described by the equation $\phi = \frac{hc}{\lambda}$. Rearranging this relationship to solve for the wavelength gives $\lambda = \frac{hc}{\phi}$. Using the constants Planck's constant $h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$ and the speed of light $c = 3 \times 10^8 \text{ m/s}$, and converting the work function of 4.0 eV to joules by multiplying by the elementary charge $1.6 \times 10^{-19} \text{ C}$, the calculation is:

$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.0 \times 1.6 \times 10^{-19}} \approx 3.107 \times 10^{-7} \text{ m}$

Converting this value from meters to nanometers results in a wavelength of approximately 310 nm, which is the limit for photoelectric emission from the given substance.

Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons, $K.E.$, to the frequency of incident light, $\nu$, through the linear expression $K.E. = h\nu - \phi$, where $h$ is Planck's constant and $\phi$ represents the work function of the material. When plotting $K.E.$ on the vertical axis against frequency on the horizontal axis, this equation takes the form of a straight line $y = mx + c$, where the slope $m$ corresponds directly to Planck's constant. Because Planck's constant is a universal physical value, the resulting slope is identical for all metals and remains completely independent of the intensity of the incident radiation.

Einstein's photoelectric equation dictates that the energy of an incident photon $hf$ is equal to the sum of the material's work function $W$ and the maximum kinetic energy of the ejected electron $\frac{1}{2}mv^2$. For the two scenarios provided, we can write the energy conservation as $hf_1 = W + \frac{1}{2}mv_1^2$ and $hf_2 = W + \frac{1}{2}mv_2^2$. By subtracting the second equation from the first, the work function terms cancel out, simplifying the relationship to $h(f_1 - f_2) = \frac{1}{2}m(v_1^2 - v_2^2)$. Rearranging this equality to isolate the velocity terms results in the expression
$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$

The work function of a metal denotes the minimum energy required to eject an electron, and this energy is inversely related to the threshold wavelength through the relation $\lambda = \frac{hc}{\Phi}$. Consequently, the ratio of the threshold wavelengths for sodium and copper follows the inverse proportion of their respective work functions, expressed as $\frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{\Phi_{Cu}}{\Phi_{Na}}$. Substituting the given values of 4.5 eV for copper and 2.3 eV for sodium results in $\frac{4.5}{2.3}$, which simplifies to approximately 2, yielding a ratio of $2:1$.

Within the framework of molecular orbital theory, covalent bond formation is fundamentally understood through the wave nature of electrons. As atomic centers approach each other, their respective electron wave functions overlap and interfere to create a new molecular orbital, mathematically represented as $\psi_{MO} = c_1\psi_1 \pm c_2\psi_2$. This wave-like superposition of orbitals is the essential mechanism that concentrates electron density between nuclei, ultimately stabilizing the connection between atoms.

When an electron moves through a uniform electric field, it experiences a constant force that directly affects its velocity. Because the electron carries a charge of $-e$ and is subjected to an electric field of $\vec{E} = -2 E_0 \hat{i}$, it experiences an electrostatic force of $\vec{F} = q\vec{E} = (-e)(-2 E_0 \hat{i}) = 2 e E_0 \hat{i}$. Applying Newton's second law, this force results in a constant acceleration of $a = \frac{2 e E_0}{m}$ in the same direction as its initial motion.

The instantaneous velocity $v(t)$ at any time $t$ is determined by the kinematic relation $v(t) = v_0 + at$, which becomes $v(t) = v_0 + \left( \frac{2 e E_0}{m} \right) t$. Since the de Broglie wavelength is inversely proportional to momentum, expressed as $\lambda = \frac{h}{m v(t)}$, substituting the expression for velocity gives $\lambda(t) = \frac{h}{m \left( v_0 + \frac{2 e E_0 t}{m} \right)}$. By using the given initial wavelength condition $\lambda_0 = \frac{h}{4 m v_0}$, which implies $h = 4 \lambda_0 m v_0$, we substitute this value into the equation to get $\lambda(t) = \frac{4 \lambda_0 m v_0}{m \left( v_0 + \frac{2 e E_0 t}{m} \right)}$. Dividing both the numerator and the denominator by $v_0$ yields the expression $\lambda(t) = \frac{4 \lambda_0}{1 + \frac{2 E_0 e t}{m v_0}}$.

The de Broglie wavelength for a charged particle accelerated through a potential difference $V$ is defined by the relation $\lambda = \frac{h}{\sqrt{2meV}}$, where $h$ is Planck's constant, $m$ is the mass of the particle, and $e$ is its charge. Since both the proton and the deuteron possess the same elementary charge $e$, the relationship simplifies such that wavelength is inversely proportional to the square root of the product of mass and potential, or $\lambda \propto \frac{1}{\sqrt{mV}}$. Recognizing that a deuteron is twice as massive as a proton, with $m_d = 2m_p$, we can determine the ratio of their wavelengths as $\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d V_d}{m_p V_p}} = \sqrt{\frac{2m_p V_d}{m_p V_p}} = \sqrt{2 \frac{V_d}{V_p}}$. Given that the ratio of the wavelengths is $\frac{1}{\sqrt{2}}$, we substitute this value to obtain $\frac{1}{\sqrt{2}} = \sqrt{2 \frac{V_d}{V_p}}$. Squaring both sides of this equation results in $\frac{1}{2} = 2 \frac{V_d}{V_p}$, which leads to the final ratio of the accelerating potentials, $\frac{V_p}{V_d} = 4$.

The de Broglie wavelength of a particle is fundamentally determined by its momentum, expressed as $\lambda = \frac{h}{mv}$. An electron with initial velocity $v_0$ moving through a constant electric field $\vec{E} = -E_0 \hat{i}$ experiences an electrostatic force $F = eE_0$ that causes it to accelerate at a rate of $a = \frac{eE_0}{m}$.

As time $t$ progresses, the velocity of the electron increases to $v = v_0 + \frac{eE_0}{m}t$. Substituting this into the wavelength formula, we get $\lambda = \frac{h}{m(v_0 + \frac{eE_0}{m}t)}$. By factoring $v_0$ out from the denominator, the expression becomes $\lambda = \frac{h}{mv_0(1 + \frac{eE_0 t}{m v_0})}$. Given that the initial wavelength is $\lambda_0 = \frac{h}{mv_0}$, the wavelength at time $t$ simplifies to $\lambda = \frac{\lambda_0}{1 + \frac{eE_0 t}{m v_0}}$.