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(C) For a particle of mass $m$ confined to a one-dimensional box of length $L$,the energy levels are given by $E_k = \frac{k^2 h^2}{8 m L^2}$.Given $L

Vedclass · Accepted Answer

$n=6, m=4$

Vedclass · Answer

(C) For a particle of mass $m$ confined to a one-dimensional box of length $L$,the energy levels are given by $E_k = \frac{k^2 h^2}{8 m L^2}$.Given $L = \sqrt{\frac{35 h \lambda}{8 m c}}$,we have $L^2 = \frac{35 h \lambda}{8 m c}$.Substituting $L^2$ into the energy expression:$E_k = \frac{k^2 h^2}{8 m (35 h \lambda / 8 m c)} = \frac{k^2 h c}{35 \lambda}$.When the electron absorbs a photon of wavelength $\lambda$,it transitions from $k=1$ to $n$:$E_n - E_1 = \frac{h c}{\lambda}$$\frac{n^2 h c}{35 \lambda} - \frac{1^2 h c}{35 \lambda} = \frac{h c}{\lambda}$$\frac{n^2 - 1}{35} = 1 \Rightarrow n^2 - 1 = 35 \Rightarrow n^2 = 36 \Rightarrow n = 6$.Next,the electron transitions from $n=6$ to $m$ by emitting a photon of wavelength $\lambda^{\prime} = 1.75 \lambda = \frac{7}{4} \lambda$:$E_6 - E_m = \frac{h c}{\lambda^{\prime}}$$\frac{6^2 h c}{35 \lambda} - \frac{m^2 h c}{35 \lambda} = \frac{h c}{1.75 \lambda}$$\frac{36 - m^2}{35} = \frac{1}{1.75} = \frac{1}{7/4} = \frac{4}{7}$$36 - m^2 = 35 	imes \frac{4}{7} = 5 	imes 4 = 20$$m^2 = 36 - 20 = 16 \Rightarrow m = 4$.Thus,$n=6$ and $m=4$.

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