Question
Energy of an electron is given by $$E = - 2.178 \times {10^{ - 18}}J\left( {\frac{{{Z^2}}}{{{n^2}}}} \right).$$
Wavelength of light required to excite an electron in an hydrogen atom from level $$n = 1$$ to $$n = 2$$ will be:
$$\left( {h = 6.62 \times {{10}^{ - 34}}Js\,{\text{and}}\,c = 3.0 \times {{10}^8}m{s^{ - 1}}} \right)$$
A.
$$1.214 \times {10^{ - 7}}m$$
B.
$$2.816 \times {10^{ - 7}}m$$
C.
$$6.500 \times {10^{ - 7}}m$$
D.
$$8.500 \times {10^{ - 7}}m$$
Answer :
$$1.214 \times {10^{ - 7}}m$$
Solution :
$$\eqalign{
& \Delta E = 2.178 \times {10^{ - 18}}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{{hc}}{\lambda } \cr
& 2.17 \times {10^{ - 18}} \times \frac{3}{4} = \frac{{hc}}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } \cr
& \lambda = \frac{{6.62 \times \times {{10}^{ - 34}}}}{{2.17 \times {{10}^{ - 18}}}}\frac{{3 \times {{10}^8} \times 4}}{{ \times 3}} \cr
& = 1.214 \times {10^{ - 7}}m \cr} $$