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The recoil speed of a hydrogen atom af...

The recoil speed of a hydrogen atom after it emits a photon is going form n=5 state to n =1 state is ….. m/s.

A

`10^(-4) m//s`

B

`2 xx 10^(-2) m//s`

C

`4.2 m//s`

D

`3.8 xx 10^(-2) m//s`

Text Solution

Verified by Experts

The correct Answer is:
C
(a) `DeltaE=13.6 (1/1^(2)-1/5^(2))=(hc)/lambda+1/2 mv^(2)`
(b) `0=mv-h/lambda`
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The ionization energy of hydrogen atom in the ground state is 1312 kJ "mol"^(-1) . Calculate the wavelength of radiation emitted when the electron in hydrogen atom makes a transition from n = 2 state to n = 1 state (Planck’s constant, h = 6.626 xx 10^(-34) Js , velocity of light, c = 3 xx 10^8 m s^(-1) , Avogadro’s constant, N_A = 6.0237 xx 10^23 "mol"^(-1) ).

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Knowledge Check

  • Energy of a hydrogen atom with principal quantum number n is shown by E = (-13.6)/(n^(2)) eV . The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately.

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    C
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    D
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    A
    `A_(n)~~A+(B)/(lamda_(n)^(2))`
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