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  • If the wavelength for an electron emitted from -atom is <img alt="3.3 \times 10^{-10} \mathrm{~m}" src

If the wavelength for an electron emitted from \mathrm{H}-atom is 3.3 \times 10^{-10} \mathrm{~m}, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is________ times. (Nearest integer)
[Given : \mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} ]
Mass of electron =9.1 \times 10^{-31} \mathrm{~kg}

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\lambda = 3.3\times 10^{-10}m}
\mathrm{h = 6.626\times 10^{-34}\, J\, s}
\mathrm{me= 9.1\times 10^{-31}}

\mathrm{\lambda = \frac{h}{\sqrt{2mk}}}
\mathrm{k= \frac{h^{2}}{2m\lambda ^{2}}= \frac{\left ( 6.6\times 10^{-34} \right )^{2}}{2\times 9.1\times 10^{-31}\times \left ( 3.3\times 10^{-10} \right )^{2}}}
\mathrm{k= \frac{43.9\times 10^{-68}}{2\times 9.1\times 10^{-31}\times 10.89\times 10^{-20}}}
\mathrm{k= 2.215\times 10^{-18}}

Energy absorbed -
\mathrm{Eabs= Ereq+k---(1)}

devide eqn (1) by Ereq. we get -
\mathrm{\frac{Eabs}{Ereq}= \frac{Ereq}{Ereq}+\frac{k}{Ereq}}
             \mathrm{= 1+\frac{k}{Ereq}= 1+\frac{2.215\times 10^{-18}}{13.6\times 1.602\times 10^{-19}}= 2.0116}

Hence the answer is 2.  

 

Posted by

rishi.raj

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