JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Atomic Structure

The surface of sodium metal is irradiated with radiation of wavelength x nm . The kinetic energy of ejected electrons is

2.8 \times 1 0^{- 20} J

. The work function of sodium is 2.3 eV . The value of x is
____

\times 1 0^{2} nm

. (Nearest integer)
(Given :

h = 6.6 \times 1 0^{- 34} J s; 1 eV = 1.60 \times 1 0^{- 19} J

;

c = 3.0 \times 1 0^{8} ms^{- 1}

)

Answer: 5

Solution

The photon energy divides between the work function and the kinetic energy of the ejected electron,

\frac{h c}{λ} = W + KE

. With

W = 2.3 \times 1.60 \times 1 0^{- 19} = 3.68 \times 1 0^{- 19}

J and

KE = 2.8 \times 1 0^{- 20}

J, the photon energy is

3.96 \times 1 0^{- 19}

J. Then

λ = \frac{h c}{3.96 \times 1 0 ^{- 19}} = \frac{6.6 \times 1 0 ^{- 34} \times 3.0 \times 1 0 ^{8}}{3.96 \times 1 0 ^{- 19}} = 5.0 \times 1 0^{- 7} m = 500 nm .

Written as

x \times 1 0^{2}

nm, the value of x is

5

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Integer answer