Answer:
Energy of photon, \[hv\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[\therefore \] \[\frac{{{(hv)}_{2\to 1}}}{{{(hv)}_{\infty \to 1}}}=\frac{\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)}{\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)}=\frac{3}{4}=\mathbf{3:4}.\]
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