JEE MAIN - Physics (2018 (Offline) - No. 27)
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 nth state and the ground state respectively. Let
$${\Lambda _n}$$ be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)
$${\Lambda _n} \approx A + {B \over {\lambda _n^2}}$$
$${\Lambda _n} \approx A + B{\lambda _n}$$
$$\Lambda _n^2 \approx A + B\lambda _n^2$$
$$\Lambda _n^2 \approx \lambda$$
Explanation
We know,
Wavelength of emitted photon from n2 state to n1 state is
$${1 \over \lambda }$$ = RZ2 $$\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
Here electron comes from nth state to ground state (n = 1),
then the wavelength of photon is ,
$${1 \over {{\Lambda _n}}}$$ = RZ2 $$\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$n = $${1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} \right)^{ - 1}}$$
As n is very large, so using binomial theorem
$$\Lambda $$n = $${1 \over {R{Z^2}}}\left( {1 + {1 \over {{n^2}}}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$n = $${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{n^2}}}} \right)$$
We know,
$$\lambda $$n = $${{2\pi r} \over n}$$
= 2$$\pi $$ $${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}$$
$$\therefore\,\,\,$$ $$\lambda $$n $$ \propto $$ n
$$ \Rightarrow $$$$\,\,\,$$ n = K $$\lambda $$n
$$\therefore\,\,\,$$ $$\Lambda $$n = $${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} \right)}^2}}}} \right)$$
Let A = $${1 \over {R{Z^2}}}$$ and B = $${1 \over {{K^2}R{Z^2}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$n = A + $${B \over {\lambda _n^2}}$$
Wavelength of emitted photon from n2 state to n1 state is
$${1 \over \lambda }$$ = RZ2 $$\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
Here electron comes from nth state to ground state (n = 1),
then the wavelength of photon is ,
$${1 \over {{\Lambda _n}}}$$ = RZ2 $$\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$n = $${1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} \right)^{ - 1}}$$
As n is very large, so using binomial theorem
$$\Lambda $$n = $${1 \over {R{Z^2}}}\left( {1 + {1 \over {{n^2}}}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$n = $${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{n^2}}}} \right)$$
We know,
$$\lambda $$n = $${{2\pi r} \over n}$$
= 2$$\pi $$ $${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}$$
$$\therefore\,\,\,$$ $$\lambda $$n $$ \propto $$ n
$$ \Rightarrow $$$$\,\,\,$$ n = K $$\lambda $$n
$$\therefore\,\,\,$$ $$\Lambda $$n = $${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} \right)}^2}}}} \right)$$
Let A = $${1 \over {R{Z^2}}}$$ and B = $${1 \over {{K^2}R{Z^2}}}$$
$$ \Rightarrow $$$$\,\,\,$$ $$\Lambda $$n = A + $${B \over {\lambda _n^2}}$$
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