Question

Calculate the wavelength associated with an electron of mass 9.1  10-31 kg

moving with the velocity of 103 m/sec. (h = 6.6  10-34 kg m2 s

-1

)​

Answer 1

Answer:

The concept of De-Broglie Wavelength would be used to solve this problem since the wavelength of electron is a microscopic quantity.

According to De-Broglie,

$\lambda = \dfrac{h}{\rho}$

Here, 'λ' refers to the De-Broglie Wavelength, 'h' refers to Planck's constant and 'ρ' refers to momentum of the particle.

According to the question,

• h = 6.63 × 10⁻³⁴ Kg.m²/s
• m = 9.1 × 10⁻³¹ Kg
• v = 10³ m/s

Calculating momentum of the electron we get:

→ ρ = m × v

→ ρ = 9.1 × 10⁻³¹ Kg × 10³ m/s

→ ρ = 9.1 × 10⁻²⁸ Kg.m/s

Substituting the given information, we get:

$\implies \lambda = \dfrac{6.63 \times 10^{-34}\:\: Kg.m^2/s}{9.1 \times 10^{-28} \:\: Kg.m/s}\\\\\\\implies \lambda = 0.728 \times 10^{-6}\:\:m \implies 728 \times 10^{-9}\:\:m$

Hence the value of wavelength of the electron is 728 n.m

Answer 2

Answer:

here is your answer

please mark as brainlist

Explanation:

the concept of de-broglie wavelength would be used to solve this problem since the wavelength of electron is a microscopic quantity.

according to de-broglie,

\lambda = \dfrac{h}{\rho}λ=

ρ

h

here, 'λ' refers to the de-broglie wavelength, 'h' refers to planck's constant and 'ρ' refers to momentum of the particle.

according to the question,

h = 6.63 × 10⁻³⁴ kg.m²/s

m = 9.1 × 10⁻³¹ kg

v = 10³ m/s

calculating momentum of the electron we get:

→ ρ = m × v

→ ρ = 9.1 × 10⁻³¹ kg × 10³ m/s

→ ρ = 9.1 × 10⁻²⁸ kg.m/s

substituting the given information, we get:

\begin{gathered}\implies \lambda = \dfrac{6.63 \times 10^{-34}\:\: kg.m^2/s}{9.1 \times 10^{-28} \:\: kg.m/s}\\\\\\\implies \lambda = 0.728 \times 10^{-6}\:\:m \implies 728 \times 10^{-9}\:\:m\end{gathered}