The surface of a metal is illuminated alternately with photons of energies $\mathrm{E}_{1}=4 \mathrm{eV}$ and $\mathrm{E}_{2}=2.5 \mathrm{eV}$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is ____.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Modern Physics

The surface of a metal is illuminated alternately with photons of energies

E_{1} = 4 eV

and

E_{2} = 2.5 eV

respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is ____.

Answer: 2

Solution

Einstein's photoelectric relation gives

E = ϕ + K

for each illumination. The maximum speeds are in the ratio

2 : 1

, so the kinetic energies are in the ratio

4 : 1

, giving

K_{1} = 4 K_{2}

. Subtracting the two cases,

E_{1} - E_{2} = K_{1} - K_{2}

, i.e.

4 - 2.5 = 4 K_{2} - K_{2}

, so

1.5 = 3 K_{2}

and

K_{2} = 0.5

eV. Substituting into

E_{2} = ϕ + K_{2}

2.5 = ϕ + 0.5

, so the work function is

ϕ = 2

eV.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Integer answer

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