KVPY Sample Paper KVPY Stream-SX Model Paper-20

An electron typically spends about \[{{10}^{-8}}\] in an excited state before it drops to a lower state by emitting a photon. Number of revolutions made by an electron in \[n=2,\] Bohr's orbit in \[1.00\times {{10}^{-\,8}}s\] is

A) about 106

B) about 1014

C) about 1016

D) about 102

Correct Answer: A

Solution :

Frequency of revolution in a Bohr's orbit is

\[fn=\frac{-{{E}_{1}}}{h}\left( \frac{2}{{{n}^{3}}} \right)\]

Where \[{{E}_{1}}\]=energy in \[n=1\]state.

So, \[{{f}_{2}}=\frac{-{{E}_{1}}}{h}\times \frac{2}{{{2}^{3}}}=0.0823\times {{10}^{15}}rps\]\[=8.23\times {{10}^{14}}\,rps\]

Number of revolutions made by electron in \[1\times {{10}^{-8}}s\]is

\[\begin{align} & N={{f}_{2}}\times \Delta t=8.23\times {{10}^{14}}\times 1\times {{10}^{-8}} \\ & =8.23\times {{10}^{6}} \\ \end{align}\]

Revolutions