Question

Radius of the fourth orbit in hydrogen atom is0.85 nm. Calculate the velocity of the electron in this orbit (mass of electron = 9.1 x 100 kg).​

Answer 1

Answer:

$\displaystyle{v=5.45\times10^{5} \ m/sec}$

Explanation:

From Bohr's postulate  

The angular momentum is given by  

$\displaystyle{mvr=\frac{nh}{2\pi}}\\\\\displaystyle{Can \ be \ write \ as}\\\\\displaystyle{v=\frac{nh}{2\pi mr}}$

We have ;  

$\displaystyle{n=4}$

$\displaystyle{m=9.1\times10^{-31} \ kg}$

$\displaystyle{r=0.85\times10^{-9}}$

Now putting in equation we get

$\displaystyle{v=\frac{4\times6.626\times10^{-34}}{2\pi \times9.1\times10^{-31}\times0.85\times10^{-9}}}$

$\displaystyle{v=5.45\times10^{5} \ m/sec}$

Thus we get answer .  

Answer 2

Solution:

(i) 9.11 × 10-28 g is the mass of 1 electron No. of electrons 1 g of mass = = 0.1099 × 1028 = 1.099 × 1027 (ii) Mass of 1 electron = 9.1 × 10-31 kg Mass of 6.023 × 1023 electrons = 6.023 × 1023 × 1.6 ... The value of ionization energy of hydrogen atom is 2.18 10—18 J (n=1) which is greater than that in the 5th orbit ...