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Question: A 20 amu atom emits photon of 6.6 Å while making a transition from excited state to ground state. Th...

A 20 amu atom emits photon of 6.6 Å while making a transition from excited state to ground state. The recoil energy of the atom will be –

A

1.5 × 10–23 J

B

3.5 × 10–23 J

C

5.1 × 10–23 J

D

7.5 × 10–23 J

Answer

1.5 × 10–23 J

Explanation

Solution

2mE\sqrt{2mE} = hλ\frac{h}{\lambda}

E =h22mλ2\frac{h^{2}}{2m\lambda^{2}}=(6.6×10−34)22×20×1.66×10−27×(6.6×10−10)2\frac{(6.6 \times 10^{- 34})^{2}}{2 \times 20 \times 1.66 \times 10^{- 27} \times (6.6 \times 10^{- 10})^{2}}

= 10−48+2740×1.66\frac{10^{- 48 + 27}}{40 \times 1.66} = 14×1.66\frac{1}{4 \times 1.66}×10–22 J

= 1.5 × 10–23 J