The ionization energy of $H$ atom is $x k J$ . The energy required for the electron to jump from $n = 2$ to $n = 3$ will be:

5 x k J

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\frac{36 x}{5} k J

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\frac{5 x}{36} k J

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\frac{9 x}{4} k J

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Solution

The correct option is D $\frac{5 x}{36} k J$
The energy of an electron in the $n^{t h}$ orbit of hydrogen atom $= \frac{- 13.6}{n^{2}} e V$ .
$I . E . = + 13.6 e V = x k J$

The energy difference of $2^{n d}$ and $3^{r d}$ orbit
$Δ E = x [\frac{1}{4} - \frac{1}{9}] k J$
$Δ E = x \times \frac{5}{36} k J$
$Δ E = \frac{5 x}{36} k J$

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The ionization energy of H atom is x kJ. The energy required for the electron to jump from n=2 to n=3 will be:

The correct option is D 5x36 kJThe energy of an electron in the nth orbit of hydrogen atom =−13.6n2 eV.I.E.=+13.6 eV=x kJThe energy difference of 2nd and 3rd orbit ΔE=x[14−19] kJΔE=x×536 kJΔE=5x36 kJ

The ionization energy of $H$ atom is $x k J$ . The energy required for the electron to jump from $n = 2$ to $n = 3$ will be: