Question

The work-function of caesium metal is $2.14eV$. 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. $V_0=0.35V,V_{\max }=350.7km{s}^{-1}$
• B. $V_0=.2V,V_{\max }=250km{s}^{-1}$
• C. $V_0=1.2V,V_{\max }=250km{s}^{-1}$
• D. None of the above

Answer 1

Answer: (A)

Given, work-function of caesium metal ${\varphi }_0=2.14eV$
Frequency of light $v=6\times 10^{4}Hz$
$K{E}_{\max }=hv-\varphi =0.35eV$
(i) Let stopping potential be $V_0$
We know that
$K{E}_{\max }=e{V}_0$
$0.35eV=e{V}_0$
$V_0=0.35V$
(ii) Maximum kinetic energy $K{E}_{\max }=\frac{1}{2}m{v}_{\max }^2$
$0.35eV=\frac{1}{2}m{v}_{\max }^2$
(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 $v_{\max }=350713.55m{s}^{-1}$
$v_{\max }=350.7km{s}^{-1}$