The value of $I E_{1}$ for $H e$ is 2372 kJ/mol, the value of $I E_{2}$ should be:

5248 kJ/mol

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

4852 kJ/mol

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

5648 kJ/mol

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

4508 kJ/mol

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

Open in App

Solution

The correct option is B 5248 kJ/mol
Since ${I E}_{2}$ is around $54$ eV (Bohr's Theory)
Or $\sim 5248$ kJ/mol (Unit conversion)
E= $13.6. \frac{Z_{e f f}^{2}}{n_{2}} (Z_{e f f} \approx 1.7) n = 1$

Suggest Corrections

Similar questions

{(I E)}_{1}

{(I E)}_{2}

{m o l}^{- 1}

The equilibrium constant for a reaction is 10. Calculate the value of ∆C^o at 300K

O H (g)

H (g)

O (g)

k J m o l^{- 1}

k J m o l^{- 1}

k J m o l^{- 1} .

O - H

Use bond energies to estimate the enthalpy of formation of HBr(g).
BE(H-H) = 436 kJ/mol
BE(Br-Br) = 192 kJ/mol
BE(H-Br) = 366 kJ/mol

Join BYJU'S Learning Program