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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