Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?
Answers
Answer:
Explanation:
The Balmer series corresponds to the wavelengths in the visible spectrum, where
- 1/λ = R(1/4 - 1/n²) n = 3,4,5, where the second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1.
Thus,
1/λ = R (1/n²1 - 1/n²2)
1/λ = 1.097 × 10`7 ( 1/2² - 1/1²)
= 1.097 × 10`7 ( 1/4-1)
λ = 4 / 1.097 × 3 × 10`7
= 1.215 × 10`7
= 122nm
Thus, the latter photon corresponds to the wavelength of 122nm.
Answer:
Explanation:
The Balmer series corresponds to the wavelengths in the visible spectrum, where
- 1/λ = R(1/4 - 1/n²) n = 3,4,5, where the second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1.
Thus,
1/λ = R (1/n²1 - 1/n²2)
1/λ = 1.097 × 10`7 ( 1/2² - 1/1²)
= 1.097 × 10`7 ( 1/4-1)
λ = 4 / 1.097 × 3 × 10`7
= 1.215 × 10`7
= 122nm
Thus, the latter photon corresponds to the wavelength of 122nm.