Question

The work function $\left(W_{0}\right)$ of $Li , K , Mg , Ag$ and Cu are $2.42,2.25,3.70,4.30$ and $4.80\, eV$ respectively. The number of metals which undergo photoelectric effect if a radiation of wavelength $540\, nm$ falls on them is $\left(1 \,eV =1.602 \times 10^{-19} \,J \right)$

• A. 4
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (C)

For any metal to show photoeiectric effect, $v >\, v _{0}$, where $v _{0}$ is threshold frequency, $v$ is the frequency of light.

Given,

$W_{0} $ of $Li =2.42\, eV =2.42 \times 1.602 \times 10^{-19} \,J $

$=3.876 \times 10^{-19}\, J $

$W_{0}$ of $K =2.25\, eV =3.604 \times 10^{-19} \,J $

$W_{0} $ of Mg $=3.70\,eV =5.927 \times 10^{-19} \,J $

$W_{0}$ of $Ag =4.30\, eV=6.888 \times 10^{-19} \,J $

$W_{0}$ of $Cu =4.80\, eV =7.689 \times 10^{-19} \,J$

Also $ W_{0}= h v_{0} $

$\because\, v_{0} $ of $ L i=W_{0} / h $

$=\frac{3.876 \times 10^{-19}}{6.626 \times 10^{-34}}=0.584 \times 10^{15} s ^{-1} $

$\therefore \,v_{0} $ of $K=\frac{W_{0}}{h}=\frac{3.604 \times 10^{-19}}{6.626 \times 10^{-34}}$

$ =0.54 \times 10^{15}\, s ^{-1} $

$\therefore \, v_{0} $ of $Mg =\frac{5.927 \times 10^{-19}}{6.626 \times 10^{-34}}=0.89 \times 10^{15} \,s ^{-1} $

$\therefore \, v_{0} $ of $Ag =\frac{6.888 \times 10^{-19}}{6.626 \times 10^{-34}}=1.03 \times 10^{15} s ^{-1} $

$ v_{0}$ of $Cu =\frac{7.689 \times 10^{-19}}{6.626 \times 10^{-34}}$

$=1.16 \times 10^{15} \,s ^{-1}$

Frequency of light can be calculated as:

$v=\frac{c}{\lambda} $

$=\frac{3 \times 10^{8}}{540 \times 10^{-9}}=0.0056 \times 10^{17} s ^{-1} $

$ \approx 0.56 \times 10^{15} s ^{-1}$

Thus, only in case of $K, \,v >\, v_{0}$ thus it will show photoelectric effect