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25 days ago

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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

1 answer:

eduard [2.5K]25 days ago

4 0

Explanation:

The following data has been provided:

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

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

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 $\nu = \frac{c}{\lambda}$

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

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

According to the De-Broglie equation $\lambda = \frac{h}{p}$

with p = $m \times \nu$

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

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$ = $3.6 \times 10^{-26} J/m$

Squaring both sides gives us:

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

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

where m = mass of the electron

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

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

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

Since K.E = $\frac{1}{2}m \nu^{2}$

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

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

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is $7.1 \times 10^{-22} J$ .

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