Question

The velocity of the photoelectron (\(in~ms^{-1}\)) emitted when a smooth surface of a metal is made to strike with a photon of wavelength \(4\times10^{-7}m\) (\(W_{0}\) of metal \(=2.13~eV\); \(1eV=1.6\times10^{-19}J\); \(h=6.6\times10^{-34}Js\); \(m_{e}=9.1\times10^{-31}kg\))

• A. \(\frac{1}{\sqrt{5}}\times10^{6}\)
• B. \(\frac{1}{\sqrt{3}}\times10^{5}\)
• C. \(\frac{1}{\sqrt{3}}\times10^{6}\)
• D. \(\frac{1}{\sqrt{7}}\times10^{5}\) \bigskip

Hint:

For photoelectric effect: \[ K_{\max}=\frac{hc}{\lambda}-W_0 \] and \[ v=\sqrt{\frac{2K_{\max}}{m}} \] Always convert eV into joules before calculating velocity.

Answer 1

The Correct Option is C

Concept: Einstein's photoelectric equation is \[ h\nu=W_0+K_{\max} \] or \[ \frac{hc}{\lambda}=W_0+\frac12 mv^2 \] The energy of the incident photon is partly used in overcoming the work function and the remainder appears as kinetic energy of the emitted electron.

Step 1: Calculate the energy of the incident photon.
Given, \[ \lambda=4\times10^{-7}m \] \[ E=\frac{hc}{\lambda} \] \[ E=\frac{(6.6\times10^{-34})(3\times10^8)} {4\times10^{-7}} \] \[ E=4.95\times10^{-19}J \] Converting into electron volt, \[ E=\frac{4.95\times10^{-19}} {1.6\times10^{-19}} \] \[ E\approx3.09\,eV \]

Step 2: Calculate maximum kinetic energy of emitted electron.
Given, \[ W_0=2.13\,eV \] Hence, \[ K_{\max}=3.09-2.13 \] \[ K_{\max}=0.96\,eV \] Converting into joule, \[ K_{\max}=0.96(1.6\times10^{-19}) \] \[ K_{\max}=1.536\times10^{-19}J \]

Step 3: Use kinetic energy formula to obtain velocity.
\[ \frac12 mv^2=1.536\times10^{-19} \] \[ v^2= \frac{2(1.536\times10^{-19})} {9.1\times10^{-31}} \] \[ v^2\approx3.37\times10^{11} \] \[ v\approx5.8\times10^5\,ms^{-1} \] \[ v=\frac{10^6}{\sqrt3}\,ms^{-1} \] Thus, \[ \boxed{v=\frac{1}{\sqrt3}\times10^6\,ms^{-1}} \] Hence the correct option is \[ \boxed{\text{(C)}} \]