Atoms & Nuclei

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143 of 239

Question 143 of 239

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn,λg{\lambda }_{n},{\lambda }_{g} be the de Broglie wavelength of the electron in the nth{n}^{th} state and the ground state respectively. Let ∧n{∧}_{n} be the wavelength of the emitted photon in the transition from the nth{n}^{th} state to the ground state. For large n, (A, B are constants)

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Answer: (B) ∧n≈A+Bλn2{∧}_{n}\approx A+\dfrac{B}{{\lambda }_{n}^{2}}

2πr=n⋋n2\pi r=n{⋋}_{n}

⋋n=2πrn{⋋}_{n}=\dfrac{2\pi r}{n}

∴r=ron2∴r={r}_{o}{n}^{2}

⋋n=2πnron2=n(2πro)…..(i){⋋}_{n}=\dfrac{2\pi }{n}{r}_{o}{n}^{2}=n\left(2\pi {r}_{o}\right)\ldots ..\left(i\right)

1∧​n=R{112−1n2}\dfrac{1}{{∧​}_{n}}=R\left\{\dfrac{1}{{1}^{2}}-\dfrac{1}{{n}^{2}}\right\}

∧​n=1R{1+1n2−1}{∧​}_{n}=\dfrac{1}{R}\left\{1+\dfrac{1}{{n}^{2}-1}\right\}

∧​n=1R{1+1n2}………..(ii){∧​}_{n}=\dfrac{1}{R}\left\{1+\dfrac{1}{{n}^{2}}\right\}\ldots \ldots \ldots ..\left(ii\right)

from (i)(i) and (ii)(ii)

∧​n=1R{1+4n2ro2⋋n2}=A+B⋋n2{∧​}_{n}=\dfrac{1}{R}\left\{1+\dfrac{4{n}^{2}{r}_{o}^{2}}{{⋋}_{n}^{2}}\right\}=A+\dfrac{B}{{⋋}_{n}^{2}}

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