If $E_{n}$ is the energy of electron in nth orbit for hydrogen like species, then which is the correct relationship ?

E_{1} o f H = \frac{1}{2} E_{2} o f H e^{+} = \frac{1}{3} E_{3} o f L i^{2 +} = \frac{1}{4} E_{4} o f B e^{3 +}

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

E_{1} (H) = E_{2} (H e^{+}) = E_{3} (L i^{2 +}) = E_{4} (B e^{3 +})

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

E_{1} (H) = 2 E_{2} (H e^{+}) = 3 E_{3} (L e^{2 +}) = 4 E_{4} (B e^{3 +})

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

No relation

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

Open in App

Solution

The correct option is B $E_{1} (H) = E_{2} (H e^{+}) = E_{3} (L i^{2 +}) = E_{4} (B e^{3 +})$
$E = - 13.6 \times \frac{Z^{2}}{n^{2}} e V E_{1} (H) = - 13.6 \times \frac{1^{2}}{1^{2}} = - 13.6 e V; E_{2} (H e^{+}) = - 13.6 \times \frac{2^{2}}{2^{2}} = - 13.6 e V E_{3} (L i^{2 +}) = - 13.6 \times \frac{3^{2}}{3^{2}} = - 13.6 e V E_{4} (B e^{3 +}) = - 13.6 \times \frac{4^{2}}{4^{2}} = - 13.6 e V ∴ E_{1} (H) = E_{2} (H e^{+}) = E_{3} (L i^{2 +}) = E_{4} (B e^{3 +})$

Suggest Corrections

Similar questions

E

n^{t h}

n^{t h}

{He}^{+}

n^{t h}

E_{n} = - \frac{313.6}{n^{2}}

E_{n}

E_{n}

n^{t h}

E_{n}

Join BYJU'S Learning Program