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  • The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to <img alt="\frac{\mathrm{h}^{2}}{x \mathrm{~m} \mathrm{a}_{0}^{2}}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bh%7D%5E%7B2%7D%7D%7Bx%20%

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to \frac{\mathrm{h}^{2}}{x \mathrm{~m} \mathrm{a}_{0}^{2}} . The value of 10 x is_________. ( a_{0} is radius of Bohr's orbit) (Nearest integer) [Given : \pi=3.14 ]
 

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best_answer

As we have learnt,

\mathrm{mvr= \frac{nh}{2\pi }}

\mathrm{\Rightarrow m^{2}v^{2}r^{2}= \frac{n^{2}h^{2}}{4\pi^{2} }}

\mathrm{\Rightarrow mv^{2}= \frac{n^{2}h^{2}}{4\pi^{2}mr^{2} }}

\mathrm{\Rightarrow \frac{1}{2}mv^{2}= \frac{h^{2}n^{2}}{8\pi^{2}mr^{2} }=\frac{h^{2}n^{2}}{8\pi^{2}m}\times \frac{1}{r^2}}

\mathrm{\Rightarrow KE= \frac{1}{8}\frac{h^{2}n^{2}}{\pi^{2}m }\times \frac{Z^{2}}{n^{4}a_{0}^{2}}}

               \mathrm{= \frac{h^{2}}{ma_{0}^{2}}\times \frac{1}{8\pi ^{2}}\times \frac{Z^{2}}{n^{2}}}

\therefore x=32\pi ^{2}= 315. 5       

\therefore 10x= 3155

Hence, the correct answer is 3155

Posted by

sudhir.kumar

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