The angular momentum of an electron in an H a | vedclass

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(B) According to Bohr's postulate,angular momentum $L = \frac{nh}{2\pi}$.Given $L = \frac{h}{\pi}$,we have $\frac{nh}{2\pi} = \frac{h}{\pi}$,which giv

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$\frac{5x}{36}$

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(B) According to Bohr's postulate,angular momentum $L = \frac{nh}{2\pi}$.Given $L = \frac{h}{\pi}$,we have $\frac{nh}{2\pi} = \frac{h}{\pi}$,which gives $n = 2$.Therefore,the transition is from $n_1 = 2$ to $n_2 = 3$.The energy of an electron in the $n^{th}$ state is $E_n = -\frac{2.18 	imes 10^{-18}}{n^2} \ J = -\frac{x}{n^2} \ J$.The energy required for excitation is $\Delta E = E_{n+1} - E_n = -\frac{x}{3^2} - (-\frac{x}{2^2}) = x(\frac{1}{4} - \frac{1}{9}) = x(\frac{9-4}{36}) = \frac{5x}{36} \ J$.

The angular momentum of an electron in an $H$ atom in a particular $n$ state is $\frac{h}{\pi}$. What is the energy in $J$ required to excite the electron from this particular $n$ state to $(n+1)$ state? $(x = 2.18 \times 10^{-18} \ J)$

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