Question

An electron jumps from the 4^th orbit to the 2^nd orbit of hydrogen atom. Given the Rydberg’s constant R = 10⁷ m-1. The frequency in Hz of the emitted radiation is (c = 3x10⁸ m/s)

• A. \(\frac {9}{16}\) x 10¹⁵
• B. \(\frac {3}{16}\) x 10⁵
• C. \(\frac {3}{16}\) x 10¹⁵
• D. \(\frac {9}{16}\) x 10⁵

Answer 1

The Correct Option is A

The frequency of the emitted radiation can be determined using the Rydberg formula for the hydrogen atom: 
\(\frac {1}{λ}\) = R (\(\frac {1}{n_1^2}\) - \(\frac {1}{n_2^2}\)) 
We know that
ν = \(\frac {c}{λ}\) 
Substituting the given values into the Rydberg formula: 
\(\frac {1}{λ}\) = R (\(\frac {1}{2^2}\) - \(\frac {1}{4^2}\)) 
Simplifying:
\(\frac {1}{λ}\) = R (\(\frac {1}{4}\) - \(\frac {1}{16}\)) 
\(\frac {1}{λ}\) = R x \(\frac {3}{16}\)
Now, let's substitute the value of the Rydberg constant, R = 10⁷ m⁻¹: 
\(\frac {1}{λ}\) = (10⁷ m⁻¹) x \(\frac {3}{16}\) 
\(\frac {1}{λ}\) = \(\frac {3 \times 10^7}{16}\) m⁻¹
Taking the reciprocal of λ: 
λ = \(\frac {16}{3 \times 10^7}\) m 
Now, let's calculate the frequency ν: ν = c / λ 
ν = \(\frac {(3 \times 10^8 m/s)}{(16 / (3 \times 10^7) m)}\)
ν = \(\frac {9}{16}\) x 10¹⁵ Hz 
Therefore, the frequency of the emitted radiation is (A) \(\frac {9}{16}\) × 10¹⁵ Hz.