The radiation corresponding to $3 \rightarrow 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5 \times 10^{-4} \mathrm{~T}$. Assume that the radius of the largest circular circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$)

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Modern Physics

The radiation corresponding to

3 \to 2

transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of

5 \times 1 0^{- 4} T

. Assume that the radius of the largest circular circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron

= 9.1 \times 1 0^{- 31} kg

)

(A)0.82 eV✓ Correct
(B)1.88 eV
(C)1.36 eV
(D)0.16 eV

Answer: (A)

Solution

The

3 \to 2

transition in hydrogen releases

h ν = 13.6 (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = 13.6 \times \frac{5}{36} = 1.89 eV

The largest circular path belongs to the fastest photoelectrons, so the stated radius corresponds to the maximum kinetic energy. From

r = \frac{m v}{e B}

K_{ma x} = \frac{( e B r ) ^{2}}{2 m} = \frac{( 1.6 \times 1 0 ^{- 19} \times 5 \times 1 0 ^{- 4} \times 7 \times 1 0 ^{- 3} ) ^{2}}{2 \times 9.1 \times 1 0 ^{- 31}} = 1.72 \times 1 0^{- 19} J = 1.08 eV

Einstein's photoelectric equation then gives

ϕ = h ν - K_{ma x} = 1.89 - 1.08 = 0.82 eV

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Modern Physics

Solution

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