MHT CET · Physics · Dual Nature of Matter
In photoelectric experiment the stopping potential was measured to be \(V_1\) and \(V_2\) volt with incident light of wavelength \(\lambda\) and \(\frac{\lambda}{2}\) respectively the value of \(V_2\) is [ \((\phi=\) work function, \(e=\) electronic charge \(]\)
- A \(V_1+\frac{2 \phi}{e}\)
- B \(2 V_1+\frac{\phi}{e}\)
- C \(2 V_1-\frac{\phi}{e}\)
- D \(V_1-\frac{2 \phi}{e}\)
Answer & Solution
Correct Answer
(B) \(2 V_1+\frac{\phi}{e}\)
Step-by-step Solution
Detailed explanation
Einstein's' photoelectric equation dictates:
\(e V=\frac{h c}{\lambda}-\phi\)
For incident wavelength \(\lambda\) :
For incident wavelength \(\frac{\lambda}{2}\) :
\(\frac{h c}{\lambda / 2}=\phi+e V_2\)
On subtracting equation (2) and (1)
\(\begin{aligned} & \frac{2}{1}=\frac{\phi+e V_2}{\phi+e V_1} \\ & \Rightarrow 2 \phi+2 e V_1=\phi-e V_2 \\ & \Rightarrow V_2=\frac{\phi}{e}+2 V_1\end{aligned}\)
\(e V=\frac{h c}{\lambda}-\phi\)
For incident wavelength \(\lambda\) :
For incident wavelength \(\frac{\lambda}{2}\) :
\(\frac{h c}{\lambda / 2}=\phi+e V_2\)
On subtracting equation (2) and (1)
\(\begin{aligned} & \frac{2}{1}=\frac{\phi+e V_2}{\phi+e V_1} \\ & \Rightarrow 2 \phi+2 e V_1=\phi-e V_2 \\ & \Rightarrow V_2=\frac{\phi}{e}+2 V_1\end{aligned}\)
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