The amount of energy required to remove an electron from ground state of $H e^{+}$ ion is:

54.4

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13.6

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20.4

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34

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Solution

The correct option is A $54.4$ eV
The energy required to excite or de-excite an electron from certain orbit is
$E = 13.6 \times \frac{z^{2}}{n^{2}} e V$ $Z$ = Atomic number For Helium, it is $2$
$= 13.6 \times \frac{(2)^{2}}{(1)^{2}} e V$
$= 13.6 \times 4 e V$ $n$ =orbit Here it is in ground state. So $n = 1$
$\Rightarrow E = 54.4 e V$ .

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The amount of energy required to remove an electron from ground state of He+ ion is:

The correct option is A 54.4eVThe energy required to excite or de-excite an electron from certain orbit isE=13.6×z2n2eV Z= Atomic number For Helium, it is 2 =13.6×(2)2(1)2eV =13.6×4eV n=orbit Here it is in ground state. So n=1⇒E=54.4eV .

The amount of energy required to remove an electron from ground state of $H e^{+}$ ion is: