Question

When Ultraviolet radiation of a certain frequency falls on a potassium target, the photo electrons released can be stopped completely by a retarding potential of $0.6\, V.$ If the frequency of the radiation is increased by $10\%,$ this stopping potential rises to $0.9 \,V.$ The work function of potassium is

• A. 2.0 eV
• B. 2.4 eV
• C. 3.0 eV
• D. 2.8 eV

Answer 1

Answer: (B)

Initially let frequency is $f$
Then,$ K_{\max} =eV_{0} =Hf-\phi_{0}$
$\Rightarrow e\left(0.6\right)=hf -\phi_{0} \ldots\left(i\right)$
When frequency is increased by $10\%,$
frequency will be $f '= 11 f $ and stopping potential $= V_{0}' = 0.9\,V$
$ \Rightarrow e\left(0.9\right)=h\left(1.1f\right)-\phi_{0}\cdots\left(ii\right)$
Substituting h f from Eq. $\left(i\right)$ in Eq. $\left(ii\right)$, we get
$e\left(0.9\right)=1.1\left(e\left(0.6\right)+\phi_{0}\right)-\phi_{0} $
$\Rightarrow e\left(0.9\right)=e\left(0.66\right)+0.1 \phi_{0}$
$\Rightarrow e\left(0.9-0.66\right)=0.1\phi_{0} $
$\Rightarrow 0.24 eV =0.1 \phi_{0}$
So, work function is $\phi_{0}=2.4eV$