What is the electron separation energy (in $e V$ ) for $B e^{3 +}$ in the first excited state.

13.6 e V

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

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

Open in App

Solution

The correct option is D $54.4 e V$
Electron separation energy is similar to ionization energy.
For $B e^{3 +}, E_{\infty} - E_{2} = + 13.6 \frac{z^{2}}{n^{2}}$
$= 13.6 \times \frac{4^{2}}{2^{2}} = 54.4 e V$

Suggest Corrections

Similar questions

e V

B e^{3 +}

B e^{3 +} ?

eV

B e^{3 +}

e V

B e^{3 +}

B e^{3 +}

Join BYJU'S Learning Program