Question

The emission spectrum of hydrogen has thee main radiations R_1, R₂, and R₃, resulting from its electron undergoing the following transitions respectively: n=5 to n=2 , n=4 to n=2, n=3 to n=2. List them in (a) increasing energy and (b) decreasing wavelenght. (c) Increasing frequency.

Answer 1

(case a )

1/lamda = Rz^2 (1/4 - 1/25)

= 1/ lamda = R * (0.21)

lamda = 4.76/R

Case b

1/lamda = Rz^2 (1/4 - 1/16)

= lamda = 5.33/R

Case c

1/lamda = Rz^2 (1/4 - 1/9)

= lamda = 7.2/R

Therefore we can see which lamda is the greatest from the three case (R=1.097*10^7 m-1)

Order of wavelength = R3>R2>R1

We know For energy

E = hc/lamda

Therefore directly proportional to 1/lamda value

SO Energy value = E1>E2>E3

For frequency

We know

E = h* frequency

Higher the energy higher the frequency

So Frequency order = f1>f2>f3