Question

If light of wavelength ${\lambda }_1$ is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is $E_1$. If wavelength of light changes to ${\lambda }_2$ then kinetic energy of electrons also change to $E_2$. Then work function of the metal is:

• A. $\frac{E_1{E}_2({\lambda }_1-{\lambda }_2)}{{\lambda }_1{\lambda }_2}$
• B. $\frac{E_1{\lambda }_1-E_2{\lambda }_2}{{\lambda }_2-{\lambda }_1}$
• C. $\frac{E_1{\lambda }_1-E_2{\lambda }_2}{({\lambda }_1-{\lambda }_2)}$
• D. $\frac{{\lambda }_1{\lambda }_2{E}_1{E}_2}{{\lambda }_2-{\lambda }_1}$

Answer 1

The correct option is B $\frac{E_1{\lambda }_1-E_2{\lambda }_2}{{\lambda }_2-{\lambda }_1}$

From photoelectric effect the relation is as follows :

If light of wavelength λ₁ is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is E₁ given as :

$\frac{hc}{{\lambda }_1}=W+E_1$

$hc=(W+E_1){\lambda }_1.............\left(1\right)$

If wavelength of light changes to λ₂ then kinetic energy of electrons also change to E₂ then

$\frac{hc}{{\lambda }_2}=W+E_2$

$hc=(W+E_2){\lambda }_2..........\left(2\right)$

Solving equation (1) and (2)

$W{\lambda }_1+E_1{\lambda }_1=W{\lambda }_2+E_2{\lambda }_2$

$W({\lambda }_1-{\lambda }_2)=E_2{\lambda }_2-E_1{\lambda }_1$

$W=\frac{E_2{\lambda }_2-E_1{\lambda }_1}{({\lambda }_1-{\lambda }_2)}$

$W=\frac{E_1{\lambda }_1-E_2{\lambda }_2}{({\lambda }_2-{\lambda }_1)}$