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 $0.6V.$ If the frequency of the radiation is increased by $10\%,$ this stopping potential rises to $0.9V.$ 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,$K_{\max }=e{V}_0=Hf-{\varphi }_0$
$\Rightarrow e\left(0.6\right)=hf-{\varphi }_0\dots \left(i\right)$
When frequency is increased by $10\%,$
frequency will be $f^{\prime }=11f$ and stopping potential $=V_0^{\prime }=0.9V$
$\Rightarrow e\left(0.9\right)=h\left(1.1f\right)-{\varphi }_0\cdots \left(ii\right)$
Substituting h f from Eq. $\left(i\right)$ in Eq. $\left(ii\right)$, we get
$e\left(0.9\right)=1.1\left(e\left(0.6\right)+{\varphi }_0\right)-{\varphi }_0$
$\Rightarrow e\left(0.9\right)=e\left(0.66\right)+0.1{\varphi }_0$
$\Rightarrow e\left(0.9-0.66\right)=0.1{\varphi }_0$
$\Rightarrow 0.24eV=0.1{\varphi }_0$
So, work function is ${\varphi }_0=2.4eV$