Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atoms is ________ Hz (nearest integer).

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Solution

Q.
Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atoms 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.) [2024]

Ans.
(656)   Velocity ( $v$ ) of electron in $n^{t h}$ orbit of single electron specie with atomic number Z is given by:

$v = 2.18 \times 10^{6} \frac{Z}{n} m / s$ ...(i)

Radius ( $r$ ) of $n^{t h}$ orbit of single electron specie with atomic number Z is given by:

$r = 0.529 \times 10^{- 10} \frac{n^{2}}{Z} m$ ...(ii)

Circumference of $n^{t h}$ orbit is integral multiple of wavelength ( $λ$ ) of electron i.e.

$2 π r = n λ$

$λ = \frac{2 π r}{n}$ ...(iii)

  Frequency (v) is related to velocity (v) and wavelength ( $λ$ ) by:

   $ν = \frac{v}{λ}$ ...(iv)

Put $λ$ from (iii) in (iv)

$ν = \frac{v}{2 π r / n}$ ...(v)

  Put $v$ and $r$ from (i) and (ii) in (iv)

   $ν = \frac{2.18 \times 10^{6} \frac{Z}{n}}{(2 π \times 0.529 \times 10^{- 10} \frac{n^{2}}{Z}) / n} = \frac{2.18 \times 10^{16} Z^{2}}{2 π \times 0.529 n^{2}}$

  For H, Z = 1 and for first orbit n = 1

   $ν = \frac{2.18 \times 10^{16} \times {(1)}^{2}}{2 π \times 0.529 \times {(1)}^{2}} = 656 \times 10^{13} Hz$

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