For a H atom,the energy transition from n = 1 | vedclass

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(B) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $H$ atom $(Z=1)$,the energy differ

Vedclass · Accepted Answer

$163.2$

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(B) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $H$ atom $(Z=1)$,the energy difference $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = 13.6 	imes 1^2 	imes (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes \frac{3}{4} = 10.2 \ eV$. For $Be^{3+}$ ion,the atomic number $Z = 4$. The energy difference for the same transition ($n=1$ to $n=2$) is $\Delta E' = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. $\Delta E' = 13.6 	imes 4^2 	imes (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes 16 	imes \frac{3}{4} = 13.6 	imes 4 	imes 3 = 163.2 \ eV$.

For a $H$ atom,the energy transition from $n = 1$ to $n = 2$ is $10.2 \ eV$. What will be the energy for the same transition in $Be^{3+}$ ion?

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