Question

Frequency of the de-Broglie wave of election in Bohr's first orbit of hydrogen atom is _____ \(\times 10 ^{13} Hz\) (nearest integer).

[Given : \(R_H\) (Rydberg constant) = \(2.18 \times 10^{-18 }J\). h (Plank's constant) = \(6.6 \times 10 ^{-34}\)J.s.]

Answer 1

Correct answer is : 661

\(\lambda = \frac {h}{mv}\)

\(\lambda = \frac {hv}{mv^2}\)

\(\frac {mv^2}{h} = \frac {v}{\lambda} = v\) (frequency)

Given \(\frac {1}{2} mv^2 = 2.18 \times 10 ^{-18}J\)

\(h = 6.6 \times 10 ^{-34}\)

\(v = \frac {4.36 \times 10 ^{-18}}{6.6 \times 10 ^{-34}} = 660.60 \times 10 ^{13} Hz\)

\(\approx 661 \times 10 ^{13} Hz\)