Question

Energy of an electron is given by, $E=-2.178\times {10}^{-18}\left(\frac{Z^2}{n^2}\right)J$. Wavelength of light required to excite an electron in hydrogen atom from level $n=1$ to $n=2$ will be:
$(h=6.62\times {10}^{-34}Jsandc=3\times {10}^{8}m{s}^{-1})$

• A. $6.500\times {10}^{-7}m$
• B. $8.500\times {10}^{-7}m$
• C. $1.214\times {10}^{-7}m$
• D. $2.816\times {10}^{-7}m$

Answer 1

The correct option is C $1.214\times {10}^{-7}m$

$Explanation:$

Given

$E=-2.178\times {10}^{-18}J\left[\frac{Z^2}{n^2}\right]\to \left(1\right)$

For hydrogen atom $Z=1$

$\Delta E=E_2-E_1\to \left(2\right)$

$E_1=-2.178\times {10}^{-18}\left[\frac{1}{1^2}\right]\to \left(3\right)$

$E_2=-2.178\times {10}^{-18}\left[\frac{1}{2^2}\right]\to \left(4\right)$

Substitute equation $\left(3\right)$ and $\left(4\right)$ value in equation$\left(2\right)$

$\Delta E=E_2-E_1=(-2.178\times {10}^{-18}[\frac{1}{2^2}\left]\right)-(-2.178\times {10}^{-18}[\frac{1}{1^2}\left]\right)=-2.178\times {10}^{-18}[\frac{1}{2^2}-\frac{1}{1^2}]$

$\Delta E=E_2-E_1=-2.178\times {10}^{-18}[\frac{1}{2^2}-\frac{1}{1^2}]\to \left(5\right)$

$\Delta E=\frac{hc}{\lambda }\to \left(6\right)$

where,

$h=$Planck's constant$=6.62\times {10}^{-34}$

$c=$Velocity of light$=3\times {10}^8$

$\lambda =$wavelength$=?$

Equate equation $\left(5\right)$ and $\left(6\right)$, then solve them to find the wavelength.

$\Delta E=E_2-E_1=\frac{hc}{\lambda }$

$-2.178\times {10}^{-18}[\frac{1}{2^2}-\frac{1}{1^2}]=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{\lambda }$

$\Rightarrow 1.6335\times {10}^{-18}=\frac{1.986\times {10}^{-25}}{\lambda }$

$\lambda =\frac{1.986\times {10}^{-25}}{1.6335\times {10}^{-18}}\approx 1.214\times {10}^{-7}m$

$\therefore \lambda =1.214\times {10}^{-7}m$

$Hencethecorrectanswerisoption\left(C\right).$

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