JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Modern Physics

Surface of certain metal is first illuminated with light of wavelength

λ_{1} = 350 nm

and then, the light of wavelength

λ_{2} = 540 nm

. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to: (Energy of photon

= \frac{1240}{λ ( in nm )} eV

)

(A)1.8✓ Correct
(B)2.5
(C)5.6
(D)1.4

Answer: (A)

Solution

Einstein's equation says the kinetic energy of the fastest photoelectrons is

K E = \frac{1240}{λ} - ϕ

. Because the two speeds differ by a factor of

2

, the kinetic energies differ by a factor of

4

: if

K E_{2} = x

for

540

nm, then

K E_{1} = 4 x

for

350

nm. Subtracting the two equations eliminates

ϕ

\frac{1240}{350} - \frac{1240}{540} = 3 x

, giving

3.54 - 2.30 = 3 x

, so

x \approx 0.42

eV. Then

ϕ = \frac{1240}{540} - x \approx 2.30 - 0.42 \approx 1.8

eV.

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Modern Physics

Solution

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