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QJEEAdv16Ph1aa3: JEE Advanced 2016 Physics - Class 12 JEEAdv16Ph1aa3 Solution: A hydrogen atom in its ground state is irradiated by light of…

A hydrogen atom in its ground state is irradiated by light of wavelength 970 Ao\mathop A\limits^o .

Taking hc = 1.237 ×\times 10 −- 6 eVm and the ground state energy of hydrogen atom as −- 13.6 eV, the number of lines present in the emission spectrum is

Step-by-Step Verified Solution

The energy of the incident photon of wavelength λ=970\lambda=970 Ao\mathop A\limits^o is

ΔE=hcλ=1.237×10−6970×10−10=12.75eV\Delta E=\frac{h c}{\lambda}=\frac{1.237 \times 10^{-6}}{970 \times 10^{-10}}=12.75 \mathrm{eV}

amp; Final energy of electron in nth  state =−13.6n2eVamp;⇒−13.6n2=−13.6+12.75=−0.85amp;⇒n2=16⇒n=4\begin{aligned}& \text { Final energy of electron in } n^{\text {th }} \text { state }=-\frac{13.6}{n^2} \mathrm{eV} \\\\& \Rightarrow-\frac{13.6}{n^2}=-13.6+12.75=-0.85 \\\\& \Rightarrow n^2=16 \Rightarrow n=4\end{aligned}

JEE Advanced 2016 Paper 1 Offline Physics - Atoms and Nuclei Question 42 English Explanation

The hydrogen atom can make nC2=6{ }^n \mathrm{C}_2=6 transitions while returning to ground state. Thus, the emission spectrum will have six lines.

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