Question

When a metallic surface is illuminated by a light of frequency $8\times {10}^{14}$ Hz, photoelectron of maximum energy 0.5 eV is emitted. When the same surface is illuminated by light of frequency $12\times {10}^{14}Hz$, photoelectron of maximum energy 2 eV is emitted. The work function is

• A. 0.5 eV
• B. 2.85 eV
• C. 2.5 eV
• D. 3.5 eV

Answer 1

The correct option is C 2.5 eV

The Einstein's equation for photoelectric effect is given by $E_{max}=h\nu -W$ or $h\nu =E_{max}+W$ where $E_{max}=$ maximum kinetic energy of photoelectrons , $h=$ Planck's constant, $\nu =$wavelength of incident light and $W=$ work function of metal.

In first case, $(8\times {10}^{14})h=0.5+W....\left(1\right)$

In second case, $(12\times {10}^{14})h=2+W....\left(2\right)$

$\left(2\right)/\left(1\right)\Rightarrow \frac{2+W}{0.5+W}=\frac{(12\times {10}^{14})h}{(8\times {10}^{14})h}=3/2$

or $1.5+3W=4+2W$ or $W=2.5eV$