Question

The photoelectric threshold wavelength of silver is $3250 \times {10^{ - 10}}$m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2356 \times {10^{ - 10}}$m is (Given h=$4.14 \times {10^{ - 15}}$ eVs and c=$3 \times {10^8}m{s^{ - 1}}$):
A. $\approx 0.3 \times {10^6}m{s^{ - 1}}$
B. $\approx 6 \times {10^5}m{s^{ - 1}}$
C. $\approx 6 \times {10^6}m{s^{ - 1}}$
D. $\approx 61 \times {10^3}m{s^{ - 1}}$

Answer 1

In case of photoelectric effect, the cathode is hit by a beam of photons which causes electrons in the cathode to come out of the cathode and go to anode. Potential difference is maintained between cathode and anode. The intensity of the incident light determines the current and the energy of the incident light determines the maximum energy of the electron.

Formula used:
$E = \dfrac{{hc}}{\lambda } - \phi$

Complete answer:
When radiation is hit on the cathode some of the energy of that radiation will be used by the electron to come out of that electrode which is called a work function and the remaining energy is used as a kinetic energy to reach the anode. When an electron reaches anode automatically current is generated. The amount of current produced depends on the intensity of the radiation.
We can assume that each photon will bring one electron out of the metal. Hence ‘n’ photons will bring ‘n’ electrons out of the metal.
Now some energy from the photon is used as a work function to remove electrons from the metal and the rest will be the remaining energy. Hence the maximum energy of an electron will be
$E = \dfrac{{hc}}{\lambda } - \phi$
Where $\phi$ is the work function and ‘h’ is the planck's constant and $\lambda$ is the wavelength of the light and ‘c’ is the velocity of light and ‘E’ is the maximum kinetic energy.
\eqalign{ & E = \dfrac{1}{2}{m_e}{v^2} \cr & \phi = \dfrac{{hc}}{{{\lambda _t}}} \cr}
Where ${\lambda _t}$ is threshold wavelength and ${m_e}$ is mass of electron. So
\eqalign{ & E = \dfrac{{hc}}{\lambda } - \phi \cr & \Rightarrow \dfrac{1}{2}{m_e}{v^2} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _t}}} \cr & \Rightarrow v = \sqrt {\dfrac{{2hc}}{{{m_e}}}\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _t}}}} \right]} \cr & \Rightarrow v = \sqrt {\dfrac{{2hc}}{{{m_e}}}\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _t}}}} \right]} \cr & \Rightarrow v = \sqrt {\dfrac{{2\left( {4.14 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)}}{{9.1 \times {{10}^{ - 31}} \times {{10}^{ - 10}}}}\left[ {\dfrac{1}{{2536}} - \dfrac{1}{{3250}}} \right]} \cr & \therefore v \approx 6 \times {10^5}m/s \cr}

Hence option B is the correct answer.

Note:
Work function will be constant for a metal and as wavelength of light decreases the energy of radiation increases and the maximum kinetic energy of the photoelectron emitted increases. Rather than using planck's constant and electron volt second, we can use it directly in joule second to get final value in SI units.