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  • Help me please, The acceleration of an electron in the first orbit of the hydrogen atom (n=1) is :

The acceleration of an electron in the first orbit of the hydrogen atom (n=1) is :

  • Option 1)

    \frac{h^{2}}{\pi ^{2}m^{2}r^{3}}

  • Option 2)

    \frac{h^{2}}{8\pi ^{2}m^{2}r^{3}}

  • Option 3)

    \frac{h^{2}}{4\pi ^{2}m^{2}r^{3}}

  • Option 4)

    \frac{h^{2}}{4\pi m^{2}r^{3}}

 

Answers (1)

best_answer

As we learnt in

Bohr quantisation principle -

mvr=\frac{nh}{2\pi } \\2\pi r= n\lambda

- wherein

Angular momentum of an electron in  stationary orbit is quantised.

 

 Acceleration =\frac{v^{2}}{r}

\because\ \;mvr=\frac{nh}{2\pi}\ \; \;\Rightarrow\ \;v=\frac{h}{2\pi mr}(n=1)

\therefore\ \; a=\left(\frac{h}{2\pi mr} \right )^{2}.\frac{1}{r}=\frac{h^{2}}{4\pi^{2}m^{2}r^{3}}

Correct option is 3 


Option 1)

\frac{h^{2}}{\pi ^{2}m^{2}r^{3}}

This is an incorrect option.

Option 2)

\frac{h^{2}}{8\pi ^{2}m^{2}r^{3}}

This is an incorrect option.

Option 3)

\frac{h^{2}}{4\pi ^{2}m^{2}r^{3}}

This is the correct option.

Option 4)

\frac{h^{2}}{4\pi m^{2}r^{3}}

This is an incorrect option.

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