Question

the velocity of electron in third excited state of be3+

Answer 1

here's your answer..

V=(2.188×10^6) × Z/n m/s

=(2.188×10^6) × 4/4 m/s

=(2.188×10^6) m/s

Answer 2

Answer:

The velocity of electron in third excited state of $Be^{3+}$ is:

$2.18\times 10^6 m/s$.

Explanation:

Velocity of an electron in nth shell of an hydrogen like atom was given Bohr as:

$V_n=\frac{2.18\times 10^6\times Z}{n}$ (m/s)

Z = atomic number of atom

n = Number of shell

The velocity of electron in third excited state of $Be^{3+}$:

1 → 4

$V_4=\frac{2.18\times 10^6\times 4}{4} m/s=2.18\times 10^6 m/s$