Question

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0\text{V}$. This potential drops to $0.6\text{V}$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively?
$[\text{Take}\frac{hc}{e}=1.24\times {10}^{-6}{\text{JmC}}^{-1}.]$

• A. $3.78\times {10}^{–7}\text{m},1.20\text{eV}$
• B. $3.78\times {10}^{–7}\text{m},5.60\text{eV}$
• C. $1.72\times {10}^{–7}\text{m},1.20\text{eV}$
• D. $1.72\times {10}^{–7}\text{m},5.60\text{eV}$

Answer 1

The correct option is C $1.72\times {10}^{–7}\text{m},1.20\text{eV}$

As per the question given,
For first case
$\frac{hc}{\lambda }-\varphi =6\text{eV}$
For second case
$\frac{hc}{\left(4\lambda \right)}-\varphi =0.6\text{eV}$

Solving both the cases together, we get
$\frac{3hc}{4\lambda }=5.4\text{eV}$
$\Rightarrow \lambda =\frac{3hc}{4\times 5.4\text{eV}}=\frac{3\times 1.24\times {10}^{-6}}{4\times 5.4}=1.72\times {10}^{-7}\text{m}$
Putting this value in any of the above cases we get $\varphi =1.2\text{eV}$
Hence, option $\left(A\right)$ is correct.