Question

The velocity of electron in the ground state hydrogen atom is 2.18×10^6.Its velocity in the second orbit would be

Answer 1

use formula Ve= Vo × Z/ n

Vo is the given velocity and n is the orbit number, and Z is the atomic number.

Answer 2

Answer:

The velocity of the electron in the second orbit, V₂, is $1.09\times 10^{6}$ m/s.

Explanation:

Given,

The velocity of an electron in the ground state, V₁ = $2.18\times 10^{6}$ m/s.

The velocity of the electron in the second orbit, V₂ =?

Now,

• For ground-state, the orbit is first, i.e., n = $1$
• For the second orbit, n = $2$

As we know,

• The velocity of $n^{th}$ orbit can be calculated by the equation given below:
• $V_{n} = \frac{V_{1}\times Z }{n}$

Here,

• $V_{n}$ = The velocity of the $n^{th}$ orbit
• Z = The atomic number of the atom
• n = The orbit

Now, as we know

• The hydrogen's atomic number = $1$

After putting the given values in the equation, we get:

• $V_{2} = \frac{2.18 \times 10^{6} \times 1 }{2}$ = $1.09\times 10^{6}$ m/s.

Hence, the velocity of the electron in the second orbit, V₂ = $1.09\times 10^{6}$ m/s.