Question

In the early 1960s the strange optical emission lines from starlike objects that also were producing trementdous radio signals confused scientists. Finally, in 1963 Maarten Schmidt of the Mount Palomar observatory discovered that the optical spectra were just those of hydrogen, but redshifted because of the tremendous velocity of the object with respect to Earth. The object was moving away from Earth at a speed of 50,000 km/s! Compare the wavelengths of the normal and redshifted spectral lines for the Kα and Kβ lines of the hydrogen atom.

Answer 1

Step-by-step explanation:

k -alpha line corresponds to n = 2 to n = 1 transition

energy difference

= 13.6 - 13.6 / 4

= 10.2 eV

= 12375 / 10.2 A

= 1213 A

k -beta line corresponds to n = 3 to n = 1 transition

energy difference

= 13.6 - 13.6 / 9

= 12.08 eV

= 12375 / 12.08 A

= 1024 A

For red shift the formula is

v / c = Δλ / λ

40000 x 10³ / 3 x 10⁸ =Δλ / λ

1 / 6 = Δλ / 1213

Δλ = 1213/6

= 202.167 A

Wavelength of spectral line

= 1213 + 202.16

= 1415.16 A after red shift .

For k - beta line

1 / 6 = Δλ / 1024

Δλ = 1024 / 6

= 170.67 A

Wavelength of red shifted line

= 1024 + 170.67

= 1194.67 A