JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Modern Physics

A metal surface is illuminated by light of two different wavelengths

248 nm

and

310 nm

. The maximum speeds of the photoelectrons corresponding to these wavelengths are

u_{1}

and

u_{2}

, respectively. If the ratio

u_{1} : u_{2} = 2 : 1

and

h c = 1240 e V nm

, the work function of the metal is nearly

Answer: (A)

Solution

Using the photoelectric equation:

K = \frac{h c}{λ} - ϕ

.

For wavelength 248 nm:

K_{1} = \frac{1240}{248} - ϕ = 5 - ϕ

.

For wavelength 310 nm:

K_{2} = \frac{1240}{310} - ϕ = 4 - ϕ

.

Given

\frac{u _{1}}{u _{2}} = 2

, so

\frac{K _{1}}{K _{2}} = (\frac{u _{1}}{u _{2}})^{2} = 4

.

Therefore:

\frac{5 - ϕ}{4 - ϕ} = 4

.

Solving:

5 - ϕ = 4 (4 - ϕ) = 16 - 4 ϕ

3 ϕ = 11

, so

ϕ = \frac{11}{3} \approx 3.7

eV.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option