Light of wavelength λ, strikes a photoelectric surface and electrons are ejected with an energy E. If E is to be increased to exactly twice its original value, the wavelength changes to λ' , where:
Answer & explanation
Correct answer: option 3
Energy of photoelectron
$E=\frac{hc}{λ}=\phi_0⇒\frac{hc}{λ}=E+\phi_0$ (i)
Where $\phi_0$ is the work function for the metal surface (constant).
When E' = 2E, then
$2E=\frac{hc}{λ'}-\phi_0$
$⇒\frac{hc}{λ'}=2E+\phi_0$ (ii)
Dividing eq. (i) by eq. (ii), we get
$\frac{λ'}{λ}=\frac{E+\phi_0}{2E+\phi_0}$
$\frac{λ'}{λ}=\frac{(E+\phi_0)}{2(E+\frac{\phi_0}{2})}$
$∴\frac{λ'}{λ}>\frac{1}{2}$
or $λ'>\frac{λ}{2}$
$∴λ>λ'>\frac{λ}{2}$