If the energy of incident radiation is increased by 25%. The kinetic energy of the photoelectrons emitted from the metal surface increased from 0.6 eV to 0.9 eV. The initial energy is:

1.2 eV

The correct answer is Option (2) → 1.2 eV

According to photoelectric equation,

$K.E.=E_{incident}-\phi$

$⇒K.E._{initial}=E_{initial}-\phi=0.6eV$

and,

$E_{final}=1.25×E_{initial}$

$∴K.E_{final}=E_{final}-\phi=0.9eV$

$⇒E_{initial}-phi=0.6eV$   ...(1)

$1.25×E_{initial}-phi=0.9eV$   ...(2)

Subtracting (1) from (2),

$0.25E_{initial}=0.3eV$

$E_{initial}=\frac{0.3}{0.25}=1.2eV$