n4 → n1, n4 → n2, n2 → n1.
Hence, there will be 16 electrons with $\text{m}_\text{s}=-\frac{1}{2},$
$\text{E}_4-\text{E}_1=\frac{1212}{16}-\Big(-\frac{1212}{1}\Big)$
$=-82+1312=1230\text{kJ}/ \text{ mol}^{-1}$
$\text{n}_4\rightarrow\text{n}_2$
$\Rightarrow\text{E}_4-\text{E}_2$
$=\frac{1312}{16}-\Big(-\frac{1312}{4}\Big)=\frac{-1312+5248}{16}-=\frac{3936}{16}$
$=246\text{kJ/ mol}^{-1}$
$\text{n}_2\rightarrow\text{n}_1$
$\Rightarrow\text{E}_2-\text{E}_1$
$=\frac{-1312}{4}-\Big(\frac{-1312}{1}\Big)=\frac{-1312+5248}{4}=\frac{3936}{4}$
$=984\text{kJ/ mol}^{-1}$
Since $\text{E}=\frac{\text{hc}}{\lambda},$ i.e. energy and wavelength are inversely proportionalto each other,
$\text{n}_4\xrightarrow{\ \ \ \ \ \ }\text{n}_1$ will have minimum wavelength.
$\because\text{E}=\frac{\text{hc}}{\lambda}\Rightarrow\text{E}\propto\frac{1}{\lambda}$
Number of radial nodes = (n - l - 1)
For 3s orbital, n = 3, l = 0
When 'n' is principal quantum number, ‘l’ is azimuthal quantum number
[l = 0 for s-orbital and l = 1 for p-orbital]
Hence number of radial nodes = (3 - 0 - 1) = 2
For 2p orbital, n = 2, l = 1
$\therefore$ Numbers of radial nodes = (2 - 1 - 1) = 0
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Calculate the threshold wavelength.
[Given: 1 eV = 1.6 × 10-19J]