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Atoms and Nuclei question

2012 · Q147
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Atoms and Nuclei question

2012 · Q147

NEETPhysicsAtoms and NucleiMCQ+4 / −1
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be
  1. A
    24hR25m{{24hR} \over {25m}}25m24hR​
  2. B
    25hR24m{{25hR} \over {24m}}24m25hR​
  3. C
    25m24hR{{25m} \over {24hR}}24hR25m​
  4. D
    24m25hR{{24m} \over {25hR}}25hR24m​
View written solutionFree

Correct answer: A

We know,

1λ=RZ2(1n12−1n22){1 \over \lambda } = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)λ1​=RZ2(n12​1​−n22​1​)

where R = Rydberg constant, Z = atomic number

Here, n1 = 1, n2 = 5

∴\therefore∴ 1λ=R(112−152){1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{5^2}}}} \right)λ1​=R(121​−521​) = 24R25{{24R} \over {25}}2524R​

According to conservation of linear momentum, we get

Momentum of photon = Momentum of atom

⇒\Rightarrow⇒ hλ{h \over \lambda }λh​ = mv

⇒\Rightarrow⇒ v = hmλ{h \over {m\lambda }}mλh​ = hm(24R25){h \over m}\left( {{{24R} \over {25}}} \right)mh​(2524R​) = 24hR25m{{24hR} \over {25m}}25m24hR​

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