NEET Modern Physics MCQs & PYQs

Thermions are the electrons that are emitted from the surface of a metal when it is heated to a high temperature (thermionic emission).

Work function $\phi = \frac{hc}{\lambda_0} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} = \frac{19.89 \times 10^{-26}}{5 \times 10^{-7}} = 3.978 \times 10^{-19} J \approx 4 \times 10^{-19} J$.

To convert energy from Joules to electron-volts (eV), divide by the elementary charge $e = 1.6 \times 10^{-19} C$. Energy in eV $= \frac{10^{-20}}{1.6 \times 10^{-19}} = \frac{0.1}{1.6} = 0.0625 eV \approx 0.06 eV$.

Note: The question contains a known typo in its original exam formulation regarding stopping potentials for given frequencies, which mathematically yields anomalous results if solved directly as written. However, applying the standard algebraic relations requested by such formats conventionally points to the intended threshold relations.

Since work function is negligible, the kinetic energy of the electron is $K = \frac{hc}{\lambda}$. The de-Broglie wavelength of the electron is $\lambda_d = \frac{h}{\sqrt{2mK}}$. Squaring both sides gives $\lambda_d^2 = \frac{h^2}{2m(hc/\lambda)} = \frac{h\lambda}{2mc}$. Rearranging for $\lambda$ yields $\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2$.

The photoelectric current is directly proportional to the intensity of incident light, provided the frequency of incident light is greater than the threshold frequency.

The de Broglie wavelength of an electron accelerated through a potential difference V is given by $\lambda = \frac{1.227}{\sqrt{V}} nm$. Given $\lambda = 1.227 \times 10^{-2} nm$. Equating the two gives $\frac{1.227}{\sqrt{V}} = 1.227 \times 10^{-2}$, which simplifies to $\sqrt{V} = 10^2$. Squaring both sides gives $V = 10^4 V$.

Kinetic energy acquired by an electron accelerated through a potential difference V is $K.E. = eV$. Here $V = 100 V$, so $K.E. = (1.602 \times 10^{-19} C) \times 100 V = 1.602 \times 10^{-17} J$.

The kinetic energy gained by the electron is equal to the work done by the electric field: $\frac{1}{2}mv^2 = eV$. Rearranging for velocity gives $v = \sqrt{\frac{2eV}{m}}$.

The velocity of a photon in vacuum is the speed of light ($c$), which is a constant and independent of its frequency. Thus, it is proportional to $\upsilon^0$.