Question

Based on the equation : $\Delta E=-2.0\times {10}^{-18}J(\frac{1}{n_2^2}-\frac{1}{n_1^2})$, the wavelength of the light that must be absorbed to excite hydrogen electron from level $n=1$ to level $n=2$ will be:$(h=6.625\times {10}^{-34}Js,C=3\times {10}^{8}m{s}^{-1})$

• A. $2.650\times {10}^{-7}m$
• B. $1.325\times {10}^{-7}m$
• C. $5.300\times {10}^{-10}m$
• D. $1.325\times {10}^{-10}m$

Answer 1

The correct option is B $1.325\times {10}^{-7}m$

The expression is $\Delta E=-2.0\times {10}^{-18}J(\frac{1}{n_2^2}-\frac{1}{n_1^2})=\frac{hc}{\lambda }$

Substitute values in the above expression.

$-2.0\times {10}^{-18}J(\frac{1}{2^2}-\frac{1}{1^2})=\frac{(6.625\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1})}{\lambda }$

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