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 When a hydrogen atom, initially at rest emits, a photon resulting in transition n = 5 → n = 1, its recoil speed is about
  • a)
    10-4 m/s 
  • b)
    2 × 10-2 m/s
  • c)
    4.2 m/s
  • d)
    3.8 × 10-2 m/s
Correct answer is option 'C'. Can you explain this answer?
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To n = 2, the energy of the emitted photon can be calculated using the formula:

E = hc/λ

where E is the energy of the photon, h is the Planck constant (6.626 x 10^-34 J.s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.

To determine the wavelength, we can use the Rydberg formula:

1/λ = R(1/n1^2 - 1/n2^2)

where λ is the wavelength of the photon, R is the Rydberg constant (1.097 x 10^7 m^-1), n1 is the initial energy level (n1 = 5), and n2 is the final energy level (n2 = 2).

Substituting the values, we get:

1/λ = 1.097 x 10^7 (1/5^2 - 1/2^2)

1/λ = 1.097 x 10^7 (0.04 - 0.25)

1/λ = -8.227 x 10^6

λ = -1.214 x 10^-7 m

Since the wavelength is negative, we can take the absolute value to obtain the actual wavelength:

λ = 1.214 x 10^-7 m

Now we can calculate the energy of the photon:

E = hc/λ

E = (6.626 x 10^-34 J.s) x (2.998 x 10^8 m/s) / (1.214 x 10^-7 m)

E = 1.637 x 10^-18 J

Therefore, the energy of the emitted photon is 1.637 x 10^-18 J.
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