NEET Modern Physics MCQs & PYQs

From photoelectric equation: $\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV$ and $\frac{hc}{2\lambda} = \frac{hc}{\lambda_0} + \frac{eV}{4}$. Multiplying the second equation by 4 gives $\frac{2hc}{\lambda} = \frac{4hc}{\lambda_0} + eV$. Subtracting the first equation from this result yields $\frac{hc}{\lambda} = \frac{3hc}{\lambda_0}$, which simplifies to $\lambda_0 = 3\lambda$.

Work function $W = E_1 - K_1 = 5 eV - 2 eV = 3 eV$. For the second case, maximum kinetic energy $K_2 = E_2 - W = 6 eV - 3 eV = 3 eV$. The stopping potential $V_0 = \frac{K_2}{e} = 3 V$. Since the anode must be at a negative potential relative to the cathode to repel the electrons, the potential is -3 V.

Using the photoelectric equation: $\frac{hc}{\lambda} = W + 3eV_0$ and $\frac{hc}{2\lambda} = W + eV_0$. Multiplying the second equation by 3 gives $\frac{3hc}{2\lambda} = 3W + 3eV_0$. Subtracting the first equation from this gives $\frac{hc}{2\lambda} = 2W$. Thus, $W = \frac{hc}{4\lambda}$. Since $W = \frac{hc}{\lambda_0}$, we find $\lambda_0 = 4\lambda$.

Let K be the initial kinetic energy. $\frac{hc}{\lambda} = W + K$ and $\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = W + 3K$. Multiplying the first equation by 3 gives $\frac{3hc}{\lambda} = 3W + 3K$. Subtracting the second equation from this gives $\frac{hc}{\lambda} = 2W$. Therefore, the work function is $W = \frac{hc}{2\lambda}$.

Let initial energy be $E$. We have $E = W + 0.5$. When energy is increased by 20%, new energy is $1.2E$. So, $1.2E = W + 0.8$. Substituting $E = W + 0.5$ into the second equation gives $1.2(W + 0.5) = W + 0.8$. This expands to $1.2W + 0.6 = W + 0.8$, yielding $0.2W = 0.2$, which means $W = 1.0 eV$.

Maximum kinetic energy $K_{max} = E - W$. For the first light, $K_1 = 1 - 0.5 = 0.5 eV$. For the second light, $K_2 = 2.5 - 0.5 = 2.0 eV$. The ratio of their kinetic energies is $K_1/K_2 = 1/4$. The ratio of maximum speeds is $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.

Photoelectric emission takes place only when the frequency of the incident light is greater than a characteristic minimum frequency for the given metal, known as the threshold frequency.

The number of emitted photoelectrons per second is directly proportional to the intensity of the incident radiation, so it becomes 2N. However, the maximum kinetic energy depends only on the frequency of the incident radiation and the material's work function, so it remains T.

The energy of incident photons is $E = \frac{1240 eV nm}{\lambda(nm)} = \frac{1240}{200} = 6.2 eV$. The maximum kinetic energy of emitted photoelectrons is $K_{max} = E - W = 6.2 - 5.01 = 1.19 eV \approx 1.2 eV$. Therefore, the stopping potential difference magnitude is $1.2 V$ (Often referenced as $1.2 V$ depending on convention for potential difference).

Power is given by $P = \frac{n h c}{\lambda}$, where n is the number of photons emitted per second. Thus, $\frac{P_2}{P_1} = \frac{n_2 / \lambda_2}{n_1 / \lambda_1} = \left(\frac{n_2}{n_1}\right) \left(\frac{\lambda_1}{\lambda_2}\right) = \left(\frac{1.02 \times 10^{15}}{10^{15}}\right) \left(\frac{5000}{5100}\right) = 1.02 \times \frac{50}{51} = 1.00$.