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:

λ' is greater than λ/ 2 but less then λ

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 $λ'>\frac{λ}{2}$

$∴λ>λ'>\frac{λ}{2}$