NEET Modern Physics MCQs & PYQs

Energy of incident photon is $E = \frac{12400}{3000} eV = 4.13 eV$. Maximum kinetic energy is $K_{max} = E - \phi = 4.13 eV - 1 eV = 3.13 eV$. In joules, $K_{max} = 3.13 \times 1.6 \times 10^{-19} J \approx 5 \times 10^{-19} J$. Using $K_{max} = \frac{1}{2}mv^2$, we get $v = \sqrt{\frac{2 \times 5 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 10^6 m/s$.

Energy of one photon is $E = h\nu = 6.63 \times 10^{-34} \times 880 \times 10^3 = 5.834 \times 10^{-28} J$. Number of photons emitted per second is $n = \frac{P}{E} = \frac{10 \times 10^3}{5.834 \times 10^{-28}} \approx 1.71 \times 10^{31}$.

X-ray emission involves the release of high-energy electromagnetic radiation (photons) when fast-moving electrons strike a heavy target, not the emission of electrons. The other processes all involve electron emission.

From Einstein's photoelectric equation, $K_{max} = E - \phi = 6.2 eV - 4.2 eV = 2.0 eV$. Converting this energy to Joules: $2.0 eV = 2.0 \times 1.6 \times 10^{-19} J = 3.2 \times 10^{-19} J$.

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.