Atomic Structure Result Question 16-1
16. Energy of an electron is given by
$E=-2.178 \times 10^{-18} \mathrm{~J}\left(\dfrac{Z^2}{n^2}\right)$
(2013 Main)
Wavelength of light required to excite an electron in an hydrogen atom from level $n=1$ to $n=2$ will be
$\left(h=6.62 \times 10^{-34} \mathrm{Js}\right.$ and $c=3.0 \times 10^8 \mathrm{~ms}^{-1}$ )
(a) $1.214 \times 10^{-7} \mathrm{~m}$
(b) $2.816 \times 10^{-7} \mathrm{~m}$
(c) $6.500 \times 10^{-7} \mathrm{~m}$
(d) $8.500 \times 10^{-7} \mathrm{~m}$
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Answer:
Correct Answer: 16. ( a )
Solution:
- Given, in the question $E=-2.178 \times 10^{-18} \mathrm{~J}\left[\dfrac{Z^2}{n^2}\right]$
For hydrogen $\quad Z=1$,
So,
$\begin{aligned} & E_1=-2.178 \times 10^{-18} \mathrm{~J}\left[\dfrac{1}{1^2}\right] \\ & E_2=-2.178 \times 10^{-18} \mathrm{~J}\left[\dfrac{1}{2^2}\right] \end{aligned}$
Now, $E_1-E_2$ i.e.
$\Delta E=2.178 \times 10^{-18}\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)=\dfrac{h c}{\lambda}$
$\begin{aligned} & 2.178 \times 10^{-18}\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)=\dfrac{6.62 \times 10^{-34} \times 3.0 \times 10^8}{\lambda} \\ & \therefore \quad \lambda \approx 1.21 \times 10^{-7} \mathrm{~m} \end{aligned}$
BETA
