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Hydrogen atom from excited state comes t...

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength `lambda`. If R is the Rydberg cosntant, the principal quantum number n of the excited state is

A

`sqrt((lambdaR)/(lambdaR-1))`

B

`sqrt((lambda)/(lambdaR-1))`

C

`sqrt((lambdaR^(2))/(lambda R-1))`

D

`sqrt((lambdaR)/(lambda-1))`

Text Solution

Verified by Experts

The correct Answer is:
A
According to Rydberg.s formula
`1/lambda =R (1/n_(f)^(2)-1/n_(i)^(2))`
Here, `n_(f)=1, n_(i)=n`
`1/lambda=R (1l^(2)-1/n^(2)) rArr 1/lambda =R (1-1/n^(2)) ........(i)`
Multiplying equation (i) by `lambda` on both sides,
`1=lambda R(1-1/n^(2)) rArr 1/(lambda R)=1-1/n^(2)`
`rArr 1/n^(2)=1-(1)/(lambdaR) rArr 1/n^(2)=(lambdaR-1)/(lambdaR) rArr n= sqrt((lambdaR)/(lambdaR-1))`
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Knowledge Check

  • Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength lamda . If R is the Rydberg constant the principal quantum a number n of the excited state is

    A
    `sqrt((lamdaR)/(lamdaR-1))`
    B
    `sqrt((lamda)/(lamdaR-1))`
    C
    `sqrt((lamdaR^(2))/(lamdaR-1))`
    D
    `sqrt((lamdaR)/(lamda-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 enegy and total energy decrease
    B
    kinetic energy potential energy and total energy decreases
    C
    kinetic energy decreases, potential energy increases but total energy remains same
    D
    kinetic energy and total energy decreases but potential energy increases

  • If the hydrogen atoms are excited to states with principal quantum number n, then the number of possible emission lines is :

    A
    `n`
    B
    `n-1`
    C
    `(n(n-1))/2`
    D
    `(n(n+1))/2`

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