You visited us 1 times! Enjoying our articles? Unlock Full Access!

Alpha Decay

If the first ...

Question

If the first ionization energy of $H e$ atom is $13.6 e V$ , then the second ionization energy of $H e$ atom is :

27.2 e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40.8 e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

54.4 e V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

108.8 e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $54.4 e V$
Second ionisation energy,

$E = (13.6 \times Z^{2}) \frac{1}{n^{2}} e V$

$= 13.6 \times \frac{2^{2}}{1^{2}}$

$= 54.4 e V$

Hence, the correct option is $C$

Suggest Corrections

Similar questions

H e^{+}

L i^{2 +}

13.6 e V

H e^{+}

L i^{2 +}

13.6 e V

{H e}^{+}

13.6 e V

H e

24.6 e V

H e^{2 +}

Join BYJU'S Learning Program