An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let `lamda_(n),lamda_(g)` be the de Broglie wavelength of the electron in the `n^(th)` state and the ground state respectively. Let `A_(n)` be the wavelength of the emitted photon in the transition from the `n^(th)` state to the ground state. For large n, (A, B are constant)

Options

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

Answer

A

Explanation

`lamda_(n)=(h)/(mv)=(h)/(p)=(h)/(sqrt(2mK_(n)))impliesK_(n)=(h^(2))/(2mlamdan^(2))`
but `K_(n)=(h^(2))/(2mlamdan^(2))` and `K_(g)=(h^(2))/(2mlamdag^(2))`
`:.K_(g)-K_(n)=(h^(2))/(2m)[(1)/(lamda_(g)^(2))-(1)/(lamda_(n)^(2))]`
but `K_(n)=-E_(n)` for emission of photon and `K_(g)-E_(g)`
`:.E_(n)-E_(g)=(hc)/(lamda_(n))`
`K_(g)-K_(n)=(hc)/(A_(n))`
`A_(n)=(hc)/(k_(g)-k_(n))=(hc)/((h^(2))/(2m)[(1)/(lamda_(g)^(2))-(1)/(lamda_(n)^(2))])`
`A_(n)=(2mc)/(h[(lamda_(n)^(2)-lamda_(g)^(2))/(lamda_(n)^(2)lamda_(g)^(2))])=(2mclamda_(n)^(2)lamda_(g)^(2))/(h(lamda_(n)^(2)-lamda_(g)^(2)))`
`A_(n)=(2mclamda_(g)^(2))/(h[1-(lamda_(g)^(2))/(lamda_(n)^(2))])=(2mclamda_(g)^(2))/(h)[1-(lamda_(g)^(2))/(lamda_(n)^(2))]^(-1)`
`A_(n)=(2mclamda_(g)^(2))/(h)[1+(lamda_(g)^(2))/(lamda_(n)^(2))]`
`A_(n)=(2mclamda_(g)^(2))/(h)+(2mclamda_(g)^(4))/(hlamda_(n)^(2))`
`=A+(B)/(lamda_(n)^(2))`
where `A=(2mclamda_(g)^(2))/(h)` and `B=(2mclamda_(g)^(4))/(h)`