Question

For a photo electric experiment the relation between applied potential difference between cathode and anode (V) and the photo electric current (I) was found be as shown in fig. What is the frequency of incident radiation would be nearly $(h=6.6\times {10}^{-34}Js)$

• A. $0.436\times {10}^{18}Hz$
• B. $0.775\times {10}^{15}Hz$
• C. $0.421\times {10}^{25}Hz$
• D. $0.821\times {10}^{23}Hz$

Answer 1

The correct option is B $0.775\times {10}^{15}Hz$

For photo electric effect, $e{V}_0=hv$
$\Rightarrow v=\frac{e{V}_0}{h}=\frac{1.6\times {10}^{-19}\times 3.2}{6.6\times {10}^{-34}}=0.775\times {10}^{15}Hz\Rightarrow v=0.775\times {10}^{15}Hz$