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Question: The wavelength of radiation emitted is \(\lambda_{0}\) when an electron jumps from the third to the ...

The wavelength of radiation emitted is λ0\lambda_{0} when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

A

1625λ0\frac{16}{25}\lambda_{0}

B

2027λ0\frac{20}{27}\lambda_{0}

C

2720λ0\frac{27}{20}\lambda_{0}

D

2516λ0\frac{25}{16}\lambda_{0}

Answer

2027λ0\frac{20}{27}\lambda_{0}

Explanation

Solution

Wavelength of radiation in hydrogen atom is given by

1λ=R[1n12−1n22]\frac{1}{\lambda} = R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack

⇒ 1λ0=R[122−132]=R[14−19]=536R\frac{1}{\lambda_{0}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack = R\left\lbrack \frac{1}{4} - \frac{1}{9} \right\rbrack = \frac{5}{36}R …..(i)

and 1λ′=R[122−142]=R[14−116]=3R16\frac{1}{\lambda^{'}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack = R\left\lbrack \frac{1}{4} - \frac{1}{16} \right\rbrack = \frac{3R}{16} …..(ii)

From equation (i) and (ii) λ′λ=5R36×163R=2027\frac{\lambda^{'}}{\lambda} = \frac{5R}{36} \times \frac{16}{3R} = \frac{20}{27}

⇒ λ′=2027λ0\lambda^{'} = \frac{20}{27}\lambda_{0}