Question

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A) 2.18x10 (B) 7.84*10J (C) 8.62x10 J (D) 5.34x10 3 Answer A

Answer 1

Explanation:

The given data is as follows.

            Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

                                    E = $h \nu$

               $1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

               $\nu$ = $0.163 \times 10^{17} s^{-1}$

or,                $\nu$ = $1.63 \times 10^{16} s^{-1}$    

It is known that,        $\nu = \frac{c}{\lambda}$

                $1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$                  

                   $\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as,        p = $m \times \nu$

So,          $\lambda = \frac{h}{m \times \nu}$

            $m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$          

                             = $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

           $(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$    

                              = $12.96 \times 10^{-52}$  

               $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where,   m = mass of electron

So,           $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

                             = $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

                                   = $1.42 \times 10^{-21}$ J

Since,  K.E = $\frac{1}{2}m \nu^{2}$

                 = $\frac{1.42 \times 10^{-21} J}{2}$

                 = $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$.