JEE MAIN - Chemistry (2013 (Offline) - No. 23)
Energy of an electron is given by $$E = - 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$$. Wavelength of light required to excite an
electron in an hydrogen atom from level n = 1 to n = 2 will be
(h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
(h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
2.816 × 10−7 m
6.500 × 10−7 m
8.500 × 10−7 m
1.214 × 10−7 m
Explanation
Given $$E$$ = $$- 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$$
= $$ - 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$$
Electron in an hydrogen atom exited from level n = 1 to n = 2. So n2 = 2 and n1 = 1.
And for H, Z = 1
$$\therefore$$ $$E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)$$
= 1.6335 $$ \times $$ 10-18 J
We know $$E$$ = $${{hc} \over \lambda }$$
$$\therefore$$ $${{hc} \over \lambda }$$ = 1.6335 $$ \times $$ 10-18
then $$\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$$
$$ = 1.214 \times {10^{ - 7}}m$$
= $$ - 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$$
Electron in an hydrogen atom exited from level n = 1 to n = 2. So n2 = 2 and n1 = 1.
And for H, Z = 1
$$\therefore$$ $$E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)$$
= 1.6335 $$ \times $$ 10-18 J
We know $$E$$ = $${{hc} \over \lambda }$$
$$\therefore$$ $${{hc} \over \lambda }$$ = 1.6335 $$ \times $$ 10-18
then $$\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$$
$$ = 1.214 \times {10^{ - 7}}m$$
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