Question

The work-function of caesium metal is $2.14\, eV$. When light of frequency $6 \times 10^{14} \,Hz$ is incident on the metal surface, photoemission of electrons occurs.
What is the
(i) stopping potential and
(ii) maximum speed of the emitted photoelectrons?

• A. $V_{0}=0.35\, V , V_{\max }=350.7 \,kms ^{-1}$
• B. $V_{0}=.2\, V , V_{\max }=250\, kms ^{-1}$
• C. $V_{0}=1.2\, V , V_{\max }=250\, kms ^{-1}$
• D. None of the above

Answer 1

Answer: (A)

Given, work-function of caesium metal $\phi_{0}=2.14 \,eV$
Frequency of light $v=6 \times 10^{4} \,Hz$
$KE _{\max }=h v-\phi=0.35\, eV$
(i) Let stopping potential be $V_{0}$
We know that
$KE _{\max } =e V_{0} $
$0.35 eV =e V_{0} $
$ V_{0} =0.35\, V $
(ii) Maximum kinetic energy $KE _{\max }=\frac{1}{2} m v_{\max }^{2}$
$0.35 eV =\frac{1}{2} m v_{\max }^{2}$
(where, $v_{\max }$ is the maximum speed and $m$ is the mass of electron)
or $ \frac{0.35 \times 2 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=v_{\max }^{2}\left(\because e=1.6 \times 10^{-19}\right)$
or $v_{\max }^{2}=0.123 \times 10^{12}$
or $ v_{\max }=350713.55 \,ms ^{-1}$
$v_{\max }=350.7\, kms ^{-1}$