Question

Calculate the wavelength of a particle of mass m = 6.62 × 10⁻²⁷ kg moving with kinetic energy 7.425 × 10⁻¹³ J (h = 6.626 × 10⁻³⁴ kg m2 sec-1).

Answer 1

This question relates to the concept of De-Broglie Hypothesis, which states that matter particles also behave like waves. Thus, all matter particles possess the Wave-Particles Duality.


The mathematical form of De-Broglie Hypothesis is :

$\boxed{\lambda = \frac{h}{p}}$

Here,
$\lambda = Wavelength \\ \\ h = \text{Planck's Constant} \\ \\ p = Momentum$


We have the following data from question:

$Mass = m = 6.62 \times 10^{-27} \, \, kg \\ \\ \text{Kinetic Energy} = K = 7.425 \times 10^{-13}\, \, J \\ \\ \text{Planck's Constant} = h = 6.626 \times 10^{-34} \, \, J \, s$



Now, here we are given Kinetic Energy instead of Momentum. So we must find a relation between the two quantities.

We know that:
$p = mv \\ \\ \\ and \, \, K = \frac{1}{2}mv^2 \\ \\ \\ Consider \, \, K = \frac{1}{2}mv^2 \\ \\ \\ \text{Multiply and divide by m} \\ \\ \\ \implies K = \frac{1}{2} \times m^2v^2 \times \frac{1}{m} \\ \\ \\ \implies K = \frac{p^2}{2m} \\ \\ \\ \implies p^2 = 2mK \\ \\ \\ \implies p = \sqrt{2mK}$

Thus, now the De-Broglie Wave Equation can be written as:

$\lambda = \frac{h}{p} \\ \\ \\ \implies \boxed{\lambda = \frac{h}{\sqrt{2mK}}}$

Now, we have everything we need and we can find the wavelength.


$\lambda = \frac{h}{\sqrt{2mK}} \\ \\ \\ \implies \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2\times 6.62 \times 10^{-27}\times 7.425 \times 10^{-13}}} \\ \\ \\ \implies \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{98.307 \times 10^{-40}}} \\ \\ \\ \implies \lambda \approx \frac{6.626\times 1-^{-34}}{9.915 \times 10^{-20}} \\ \\ \\ \implies \boxed{\lambda \approx 6.683 \times 10^{-15} \, \, m}$

Thus, the De-Broglie Wavelength is approximately $6.683 \times 10^{-15} \, \, metres$


Hope it helps
Purva
Brainly Community

Answer 2

Answer:

Explanation:

6.7×10^-13