Atoms & Nuclei

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Question 122 of 239

The wavelength of the radiation emitted is λ0{\lambda }_{0} when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be 20xλ0\dfrac{20}{x}{\lambda }_{0}. The value of xx is________.

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Answer: 27

_____n=4
_____n=3
_____n=2
_____n=1
The wavelength of the light released when an electron jumps from higher energy state (n2)\left({n}_{2}\right) to lower energy state (n1)\left({n}_{1}\right) is 1λ=R(1n12−1n22)\dfrac{1}{\lambda }=R\left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right), here, λ\lambda is wavelength, RR is Rydberg Constant.
When transition is from second excited state to first excited state,
n=3→n=2n=3\rightarrow n=2
1λ0=R(122−132)...(1)\dfrac{1}{{\lambda }_{0}}=R\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{3}^{2}}\right)...\left(1\right)
When transition is from third excited state to second orbit
n=4→n=2n=4\rightarrow n=2
1(20xλ0)=R(122−142)...(2)\dfrac{1}{\left(\dfrac{20}{x}{\lambda }_{0}\right)}=R\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{4}^{2}}\right)...\left(2\right)
Dividing (2)by(1)\left(2\right)by\left(1\right)
⇒x20=122−142122−132=14−11614−19=316536=2720\Rightarrow \dfrac{x}{20}=\dfrac{\dfrac{1}{{2}^{2}}-\dfrac{1}{{4}^{2}}}{\dfrac{1}{{2}^{2}}-\dfrac{1}{{3}^{2}}}=\dfrac{\dfrac{1}{4}-\dfrac{1}{16}}{\dfrac{1}{4}-\dfrac{1}{9}}=\dfrac{\dfrac{3}{16}}{\dfrac{5}{36}}=\dfrac{27}{20}
⇒x=27\Rightarrow x=27

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