Question

caleulate the wave
longh of first: two lines of Lyman series.​

Answer 1

Explanation:

For Lyman series, n1=1.

For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n2=∞.

1/λ = R(1(n1)2−1/(n2)2) ⋅ Z2

R = Rydbergs constant (Also written is RH)

Z = atomic number.

since the electron is de-exited from 1(st) exited state (i.e n=2) to ground state (i.e n=1) for first line of Lyman series.

1/λ = R(1(1)2−1/(2)2) ⋅ Z2

Since the atomic number of Hydrogen is 1.

λ = 4/3⋅912 A.

1/R = 912 A.

λ = 121.6nm