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The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
A
n = 4 to n = 3 states
B
n = 3 to n = 2 states
C
n = 3 to n = 1 states
D
n = 2 to n = 1 states
Detailed Solution
Number of spectral lines emitted when atoms de-excite from level n: $N = \frac{n(n - 1)}{2}$
$\frac{n(n - 1)}{2} = 6$ gives $n(n - 1) = 12$, so n = 4. The atoms are excited to the n = 4 level.
Maximum wavelength corresponds to minimum energy of the photon, since $E = \frac{hc}{\lambda}$.
The energy levels get closer as n increases, so the smallest energy gap among the levels 1 to 4 is between n = 4 and n = 3.
$E_4 - E_3 = 13.6\left(\frac{1}{9} - \frac{1}{16}\right) = 0.66$ eV, which is less than $E_3 - E_2$ = 1.89 eV and $E_2 - E_1$ = 10.2 eV.
Hence the maximum wavelength corresponds to the transition from n = 4 to n = 3.
$\frac{n(n - 1)}{2} = 6$ gives $n(n - 1) = 12$, so n = 4. The atoms are excited to the n = 4 level.
Maximum wavelength corresponds to minimum energy of the photon, since $E = \frac{hc}{\lambda}$.
The energy levels get closer as n increases, so the smallest energy gap among the levels 1 to 4 is between n = 4 and n = 3.
$E_4 - E_3 = 13.6\left(\frac{1}{9} - \frac{1}{16}\right) = 0.66$ eV, which is less than $E_3 - E_2$ = 1.89 eV and $E_2 - E_1$ = 10.2 eV.
Hence the maximum wavelength corresponds to the transition from n = 4 to n = 3.