Question

In a photoemissive cell with exciting wavelength $\lambda$ , the fastest electron has a speed v. If the exciting wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electrons will be

• A. $V{\left(\frac{3}{4}\right)}^{\frac{1}{2}}$
• B. $V{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• C. $<V{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• D. $>V{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is D $>V{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

$K.E{.}_{max}=$ incident energy - work function
$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi$
Now, wavelength is changed to $\frac{3\lambda }{4}$
so, incident energy $=\frac{hc}{3\lambda /4}$
$=\frac{4hc}{3\lambda }$
so, $K.E{.}_{max}=\frac{4}{3}\frac{hc}{\lambda }-\varphi$
$K.E{.}_{max}=\frac{4}{3}\frac{hc}{\lambda }-\varphi -\frac{\varphi }{3}+\frac{\varphi }{3}$
$\frac{1}{2}m{v}_1^2=\frac{4}{3}\left(\frac{hc}{\lambda }\right)-\varphi +\frac{\varphi }{3}$
$=\frac{4}{3}\left(\frac{1}{2}\right)m{v}^2+\frac{\varphi }{3}$
$so,v_1^2>\frac{4}{3}{v}^2$
or $v_1>\sqrt{\frac{4}{3}}v$