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An electron in hydrogen atom first jumps...

An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively, Then

A

`z = 1//x`

B

`x = 9//4`

C

`y = 5//27`

D

`z = 5//27`

Text Solution

Verified by Experts

The correct Answer is:
B

`E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
For second excited state to first excited state
`E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))`
For first excited state to ground excited state
`E_(2) = (hc)/(lambda) [1-(1)/(4)] rArr (hc)/(lambda) [(3)/(4)]`
(A) `(E_(1))/(E_(2)) = (5)/(27)` , (B) `(E_(1))/(E_(2)) = ((hc)/(lambda_(1)))((lambda_(2))/(hc)) = (lambda_(2))/(lambda_(1)) = 27/5`
(C) `P prop 1/(lambda) rArr (P_(1))/(P_(2)) = 5/27`
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Knowledge Check

  • To excite the hydrogen atom from its ground state to second excited state…. eV energy is required.

    A
    3.4
    B
    12.09
    C
    10.2
    D
    13.6
  • The ratio of energies of electron in the first excited state to its second excited state in H-atom is ......

    A
    `1:4`
    B
    `4:9`
    C
    `9:4`
    D
    `4:1`
  • As an electron makes a transition from an excited state to the ground state of a hydrogen like atom/ion.

    A
    its kinetic energy increases but potential and total energy decreases
    B
    kinetic energy, potential energy and total energy decrease.
    C
    kinetic energy decreases, potential energy increases but total energy remain same.
    D
    kinetic energy and total energy decreases but potential energy increases.
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