Question

The frequency of revolution of electrons in H-atom and orbit 'N' is 8.23 x 1014 rev/s. The number of
radial and angular node in Ns are x and y respectively. The value of x-y is​

Answer 1

The frequency of revolution of electrons of an element of atomic no Z and orbit n is given as, f = 6.57 × 10^15 × Z²/n³

given, frequency of electrons in H-atom is 8.23 × 10¹⁴ Hz .

Z = 1, f = 8.23 × 10¹⁴ Hz

so, 8.23 × 10¹⁴ = 6.57 × 10^15 × (1)²/n³

⇒8.23/65.7 = 1/n³

⇒n³ = 65.7/8.23 ≈ 8

⇒n³ = (2)³

⇒n = 2

now we have to find radial node and angular node in ns i.e., 2s

n = 2, l = 0

angular node, y = l = 0

radial node, x = n - l - 1 = 2 - 0 - 1 = 1

hence, x - y = 1 - 0 = 1