If energy and angular momentum of electron in the ground state of hydrogen atom E0 and L0 respectively, then in a He+ ion, energy of the electron whose angular momentum is 3L0, will be

In the nth energy state of a hydrogen-like ion, Ln=nh2π and En=−13.6z2n2eV∴ En=−13.6 z22π Ln/h2=−13.6 h24π2z2Ln2∴E'E=z'z2.LnL'n2⇒E'E0=212132⇒E'=4E09

If energy and angular momentum of electron in the ground state of hydrogen atom E0 and L0 respectively, then in a He+ ion, energy of the electron whose angular momentum is 3L0, will be

In the nth energy state of a hydrogen-like ion, Ln=nh2π and En=−13.6z2n2eV∴ En=−13.6 z22π Ln/h2=−13.6 h24π2z2Ln2∴E'E=z'z2.LnL'n2⇒E'E0=212132⇒E'=4E09

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If energy and angular momentum of electron in the ground state of hydrogen atom E₀ and L₀ respectively, then in a He⁺ ion, energy of the electron whose angular momentum is 3L₀, will be

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$\frac{{3E}_{0}}{16}$

$\frac{{9E}_{0}}{64}$

$\frac{{27E}_{0}}{64}$

$\frac{{4E}_{0}}{9}$

answer is D.

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In the nth energy state of a hydrogen-like ion, $L_{n} = n \frac{h}{2 π} and E_{n} = - \frac{13 .6 z^{2}}{n^{2}} eV$

$∴ E_{n} = - \frac{13 .6 z^{2}}{{(2 π L_{n} / h)}^{2}} = - (\frac{13 .6 h^{2}}{4 π^{2}}) \frac{z^{2}}{L_{n}^{2}}$

$∴ \frac{E'}{E} = {(\frac{z'}{z})}^{2} . {(\frac{L_{n}}{{L'}_{n}})}^{2} \Rightarrow \frac{E'}{E_{0}} = {(\frac{2}{1})}^{2} {(\frac{1}{3})}^{2} \Rightarrow E' = \frac{4 E_{0}}{9}$

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If energy and angular momentum of electron in the ground state of hydrogen atom E0 and L0 respectively, then in a He+ ion, energy of the electron whose angular momentum is 3L0, will be

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If energy and angular momentum of electron in the ground state of hydrogen atom E₀ and L₀ respectively, then in a He⁺ ion, energy of the electron whose angular momentum is 3L₀, will be