Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Based on the equation : <img alt="\Delta E=-2.0\times 10^{-18}J\left ( \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right )" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20%5CDelta%20

Based on the equation : \Delta E=-2.0\times 10^{-18}J\left ( \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right )  the wavelength of the light that must be absorbed to excite hydrogen electron from level n=1 to level n=2 will be :

(h = 6.625x10-34 Js, C= 3 X 108 ms-1)

Option: 1

 1.325 x 10-7 m


Option: 2

 1.325 x 10-10 m


Option: 3

 2.650 x 10-7 m


Option: 4

 5.300 x 10-10 m


Answers (1)

best_answer

As discussed in 

\Delta E = -2.0\times 10^{-18}J \left ( \frac{1}{{n_{1}}^{2}}-\frac{1}{{n_{2}}^{2}} \right )

= -2.0\times 10^{-18}J \left ( \frac{1}{1}- \frac{1}{4} \right )

= -1.5\times 10^{-18}J

\therefore \frac{hc}{\lambda }= \left | E \right | = 1.5\times 10^{-18}J

\Rightarrow \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{\lambda }= 1.5\times 10^{-18}

\Rightarrow \lambda = 1.325\times 10^{-7}m

Posted by

Irshad Anwar

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2025 registration deadline extended to April 24; application fee, documents required

Latest Question