The speed of the electron (in ms -1 ) in the third orbit of hydrogen atom is approximately (mass of electron =9.1 × 10-31 kg )

Question

The speed of the electron (in ms -1 ) in the third orbit of hydrogen atom is approximately (mass of electron =9.1 × 10-31 kg ) (A) 3.6 × 105 (B) 2.18 × 106 (C) 7.26 × 105 (D) 2.18 × 105

Tardigrade Admin · Accepted Answer

Velocity of an electron can be given as follows:

Velocity (speed)

[v] = 2.18 \times 1 0^{6} \cdot \frac{Z}{n}

where,

Z =

atomic number

n =

number of orbit

For hydrogen atom,

Z = 1

\Rightarrow n = 3

Therefore,

v = 2.18 \times 1 0^{6} \cdot \frac{Z}{n} = 2.18 \times 1 0^{6} \times \frac{1}{3}

= 0.726 \times 1 0^{6} m s^{- 1}

v = 7.26 \times 1 0^{5} m s^{- 1}

Q. The speed of the electron (in $m s^{- 1}$ ) in the third orbit of hydrogen atom is approximately
(mass of electron $= 9.1 \times 1 0^{- 31} k g$ )

$3.6 \times 1 0^{5}$

$2.18 \times 1 0^{6}$

$7.26 \times 1 0^{5}$

$2.18 \times 1 0^{5}$

Q. The speed of the electron (in ms−1 ) in the third orbit of hydrogen atom is approximately (mass of electron =9.1×10−31kg )

3.6×105

2.18×106

7.26×105

2.18×105

Velocity of an electron can be given as follows: Velocity (speed) [v]=2.18×106⋅nZ​ where, Z= atomic number n= number of orbit For hydrogen atom, Z=1 ⇒n=3 Therefore, v=2.18×106⋅nZ​=2.18×106×31​ =0.726×106ms−1 v=7.26×105ms−1

Q. The speed of the electron (in $m s^{- 1}$ ) in the third orbit of hydrogen atom is approximately
(mass of electron $= 9.1 \times 1 0^{- 31} k g$ )

$3.6 \times 1 0^{5}$

$2.18 \times 1 0^{6}$

$7.26 \times 1 0^{5}$

$2.18 \times 1 0^{5}$