A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is ______.

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A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is underlinebb($\frac{32}{7}$lambda). Explanation: Given: n1 = 5, n2 = 1, Z = 1 and λ is wavelength of the wave, R is the Rydberg constant i.e. 1.097 $\times$ 107 m-1 and Z is the atomic number of the element. ni is lower energy level and nh is the higher energy level. Calculating the wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 5 to n = 1 state, `1/lambda = RZ^2(1/(n_i^2) - 1/(n_h^2)) $\Rightarrow$ lambda = RZ^2($\frac{1}{1}$^2 - $\frac{1}{5}$^2) $\Rightarrow$ lambda = $\frac{24}{25}$RZ^2 $\Rightarrow$ RZ^2 = (25lambda)/24` ...(i) Calculating the wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 5 to n = 2 state, `1/lambda^' = RZ^2(1/n_i^2 - 1/n_h^2) 1/lambda^' = RZ^2($\frac{1}{2}$^2 - $\frac{1}{5}$^2) = RZ^2 xx $\frac{21}{100}$ $\Rightarrow$ 1/lambda^' = (25lambda)/24 xx $\frac{21}{100}$ ...[From (i)] = $\frac{32}{7}$ lambda`

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