Question

what is the wavelength of balmer series if n=4 to n=2​

Answer 1

Explanation:

n=2 to n=1

We have to compare wavelength of transition in the H-spectrum with the Balmer transition n=4 to n=2 of He+ spectrum.

∵λH=λHe+

∴RHZH2[n121−n221]=RHZHe+2[221−421]

1×[n121−n221]=4×(41−161)

[n121−n221]=4×164−1

[n121−n221]=43

If n1=1, then n2=2,3,...

For first line n2=2, n1=1

[121−221]=11−41=43

Hence, transition n2=2 to n1=1 will give spectrum of the same wavelength as that of Balmer transition, n2=4 to n1=2 in He+.

Answer 2

Answer:

4858.67

Explanation:

here 1/R = 911Å

and n1= 2

n2= 4