JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Modern Physics

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A light of energy 12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is

\frac{x}{π} \times 1 0^{- 17} eVs

. The value of

x

is ____ (use

h = 4.14 \times 1 0^{- 15} eVs, c = 3 \times 1 0^{8} ms^{- 1}

Answer: 828

Solution

Absorbing

12.75 eV

raises the atom to energy

E_{n} = - 13.6 + 12.75 = - 0.85 eV

. Since

E_{n} = - \frac{13.6}{n ^{2}} eV

, this requires

n^{2} = 16

, that is

n = 4

. The angular momentum in this state is

L = \frac{nh}{2 π} = \frac{4 ( 4.14 \times 1 0 ^{- 15} )}{2 π} = \frac{8.28 \times 1 0 ^{- 15}}{π} = \frac{828}{π} \times 1 0^{- 17},

x = 828

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

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