Question 245: Atoms & Nuclei

Question

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let
${\lambda _n}$, ${\lambda _g}$ be the de Broglie wavelength of the electron in the n
th
state and the ground state respectively. Let
${\Lambda _n}$ 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 constants)

Accepted Answer

We know,
Wavelength of emitted photon from n
2
state to n
1
state is
${1 \over \lambda }$  =  RZ
2
$\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} ight)$
Here electron comes from n
th
state to ground state (n = 1),
then the wavelength of photon is ,
${1 \over {{\Lambda _n}}}$   =  RZ
2
$\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} ight)$
$ \Rightarrow $$\,\,\,$ $\Lambda $
n
=  ${1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} ight)^{ - 1}}$
As n is very large, so using binomial theorem
$\Lambda $
n
=  ${1 \over {R{Z^2}}}\left( {1 + {1 \over {{n^2}}}} ight)$
$ \Rightarrow $$\,\,\,$ $\Lambda $
n
=  ${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{n^2}}}} ight)$
We know,
$\lambda $
n
=  ${{2\pi r} \over n}$
=  2$\pi $ ${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} ight)	imes{{1 \over n}}}$
$	herefore\,\,\,$ $\lambda $
n
$ \propto $ n
$ \Rightarrow $$\,\,\,$ n  =  K $\lambda $
n
$	herefore\,\,\,$ $\Lambda $
n
= ${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} ight)}^2}}}} ight)$
Let A  =  ${1 \over {R{Z^2}}}$  and   B  = ${1 \over {{K^2}R{Z^2}}}$
$ \Rightarrow $$\,\,\,$ $\Lambda $
n
=  A + ${B \over {\lambda _n^2}}$

Answer

${\Lambda _n} \approx A + {B \over {\lambda _n^2}}$

Answer

${\Lambda _n} \approx A + B{\lambda _n}$

Answer

$\Lambda _n^2 \approx A + B\lambda _n^2$

Answer

$\Lambda _n^2 \approx \lambda$

Atoms & Nuclei

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