The energy difference between the first two levels of hydrogen atom is $10.2 e V$ . What is the corresponding energy difference for a singly ionized helium atom ?

10.2 e V

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20.4 e V

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40.8 e V

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81.6 e V

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Solution

The correct option is D $40.8 e V$
E $_{n} = 13.6 \frac{Z^{2}}{n^{2}}$
$Δ E_{h} = \frac{13.6 (1)^{^{2}}}{(1)^{2}} - \frac{13.6 (1)^{^{2}}}{(2)^{2}} = 10.2 e V$
$Δ E_{h e} = \frac{13.6 (2)^{^{2}}}{(1)^{2}} - \frac{13.6 (2)^{^{2}}}{(2)^{2}} = 4 \times Δ E_{h} = - 40.8 e V$

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The energy difference between the first two levels of hydrogen atom is 10.2eV. What is the corresponding energy difference for a singly ionized helium atom ?

The correct option is D 40.8eVEn=13.6Z2n2ΔEh=13.6(1)2(1)2−13.6(1)2(2)2=10.2 eVΔEhe=13.6(2)2(1)2−13.6(2)2(2)2=4×ΔEh=−40.8 eV

The energy difference between the first two levels of hydrogen atom is $10.2 e V$ . What is the corresponding energy difference for a singly ionized helium atom ?

The correct option is D $40.8 e V$
E $_{n} = 13.6 \frac{Z^{2}}{n^{2}}$
$Δ E_{h} = \frac{13.6 (1)^{^{2}}}{(1)^{2}} - \frac{13.6 (1)^{^{2}}}{(2)^{2}} = 10.2 e V$
$Δ E_{h e} = \frac{13.6 (2)^{^{2}}}{(1)^{2}} - \frac{13.6 (2)^{^{2}}}{(2)^{2}} = 4 \times Δ E_{h} = - 40.8 e V$