An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength \lambda , energy E = 1240\,eV \over \lambda (in\,nm) ) :

Atoms and Nuclei PYQ — JEE Main 2019 · 12 Apr · Shift 1 · Q68 with Solution

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Correct answer: C

Energy levels of hydrogen-like ion $\mathrm{He}^+$

For a hydrogen-like ion,

E_n=-\frac{13.6 Z^2}{n^2}\,\text{eV}

For $\mathrm{He}^+$ , $Z=2$ , so

E_n=-\frac{13.6\times 4}{n^2}=-\frac{54.4}{n^2}\,\text{eV}

Find energies of the emitted photons

Given,

E=\frac{1240}{\lambda(\text{nm})}\,\text{eV}

For $\lambda_1=108.5\,\text{nm}$ ,

E_1=\frac{1240}{108.5}\approx 11.43\,\text{eV}

For $\lambda_2=30.4\,\text{nm}$ ,

E_2=\frac{1240}{30.4}\approx 40.79\,\text{eV}

So total energy emitted in reaching ground state is

E_1+E_2\approx 11.43+40.79=52.22\,\text{eV}

Relate this to transition from initial state $n$ to ground state

If initial state is $n$ , final ground state is $1$ , then total emitted energy is

\Delta E=E_1-E_n=54.4\left(1-\frac{1}{n^2}\right)

Thus,

54.4\left(1-\frac{1}{n^2}\right)=52.22

Divide by $54.4$ :

1-\frac{1}{n^2}\approx \frac{52.22}{54.4}\approx 0.96

Hence,

\frac{1}{n^2}\approx 0.04=\frac{1}{25}

So,

n^2=25 \Rightarrow n=5

Check using the two successive transitions

For $n=5$ ,

E_5=-\frac{54.4}{25}=-2.176\,\text{eV}

If one photon has energy $40.8\,\text{eV}$ , this matches transition $2\to 1$ :

\Delta E_{2\to 1}=54.4\left(1-\frac14\right)=54.4\cdot \frac34=40.8\,\text{eV}

The other photon should then be from $5\to 2$ :

\Delta E_{5\to 2}=54.4\left(\frac14-\frac1{25}\right) =54.4\left(\frac{21}{100}\right)=11.424\,\text{eV}

This corresponds to

\lambda=\frac{1240}{11.424}\approx 108.5\,\text{nm}

which matches perfectly.

Therefore the initial excited state is

\boxed{n=5}

Option check

A: $n=4$ ❌
B: $n=7$ ❌
C: $n=5$ ✅
D: $n=6$ ❌

More from Atoms and Nuclei

View written solutionFree

Correct answer: C

Energy levels of hydrogen-like ion $\mathrm{He}^+$

For a hydrogen-like ion,

E_n=-\frac{13.6 Z^2}{n^2}\,\text{eV}

For $\mathrm{He}^+$ , $Z=2$ , so

E_n=-\frac{13.6\times 4}{n^2}=-\frac{54.4}{n^2}\,\text{eV}

Find energies of the emitted photons

Given,

E=\frac{1240}{\lambda(\text{nm})}\,\text{eV}

For $\lambda_1=108.5\,\text{nm}$ ,

E_1=\frac{1240}{108.5}\approx 11.43\,\text{eV}

For $\lambda_2=30.4\,\text{nm}$ ,

E_2=\frac{1240}{30.4}\approx 40.79\,\text{eV}

So total energy emitted in reaching ground state is

E_1+E_2\approx 11.43+40.79=52.22\,\text{eV}

Relate this to transition from initial state $n$ to ground state

If initial state is $n$ , final ground state is $1$ , then total emitted energy is

\Delta E=E_1-E_n=54.4\left(1-\frac{1}{n^2}\right)

Thus,

54.4\left(1-\frac{1}{n^2}\right)=52.22

Divide by $54.4$ :

1-\frac{1}{n^2}\approx \frac{52.22}{54.4}\approx 0.96

Hence,

\frac{1}{n^2}\approx 0.04=\frac{1}{25}

So,

n^2=25 \Rightarrow n=5

Check using the two successive transitions

For $n=5$ ,

E_5=-\frac{54.4}{25}=-2.176\,\text{eV}

If one photon has energy $40.8\,\text{eV}$ , this matches transition $2\to 1$ :

\Delta E_{2\to 1}=54.4\left(1-\frac14\right)=54.4\cdot \frac34=40.8\,\text{eV}

The other photon should then be from $5\to 2$ :

\Delta E_{5\to 2}=54.4\left(\frac14-\frac1{25}\right) =54.4\left(\frac{21}{100}\right)=11.424\,\text{eV}

This corresponds to

\lambda=\frac{1240}{11.424}\approx 108.5\,\text{nm}

which matches perfectly.

Therefore the initial excited state is

\boxed{n=5}

Option check

A: $n=4$ ❌
B: $n=7$ ❌
C: $n=5$ ✅
D: $n=6$ ❌

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