Question

An electron jumps from the 4th orbit to the 2nd 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

Answer: (A)

$\frac {9}{16}$ x 1015

Answer 2

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.