Question

If the velocity of electron in Bohr's first orbit is 2.19 x10@ms! The deBroglie's wavelength is
(A) 332 pm
(B) 313 pm
(C) 3.32 pm
(D) 3.13 pm​

Answer 1

Correct Question:

If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ ms⁻¹, calculate the de Broglie wavelength associated with it.

(A) 332 pm

(B) 313 pm

(C) 3.32 pm

(D) 3.13 pm

Answer :

• de Brogile wavelength, λ = 332 pm (option A ).

Step-by-step explanation:

Given that,

• Velocity of electron, v = 2.19 × 10⁶ ms⁻¹.

Also,

• Mass of electron, m = 9.11 × 10⁻³¹ kg.
• Planck's constant, h = 6.626 × 10⁻³⁴ kgm²s⁻¹.

$\hrulefill$

Using de Brogile's equation,

$\implies\rm{{\lambda}=\dfrac{h}{mv}}$

We get (substitute the given values),

$\implies\rm{\lambda=\dfrac{6.626\times{}10^{-34}\:kgm^2s^{-1}}{(9.11\times{}10^{-31}\:kg)\times{}(2.19\times{}10^6\:ms^{-1})}}$

$\implies\rm{\lambda=\dfrac{6.626\times{}10^{-9}}{9.11\times{}2.19}\:m}$

$\implies\rm{\lambda=0.332\times{}10^{-9}\:m}$

$\implies\rm{\lambda=3.32\times{}10^{-10}\:m=}\:\bold{332\:pm.}$