Question

An electron is moving at a speed of c/100. find its deBroglie wavelength and frequency. Neglect relativistic effects.

Answer 1

`\lambda=2.42* 10^(-10)\ m`

`\\u=1.24* 10^(18)\ s^(-1)`

Step-by-step explanation:

The expression for the deBroglie wavelength is:

`\lambda=\frac {h}{m* v}`

Where,

`\lambda` is the deBroglie wavelength

h is Plank's constant having value
`6.626* 10^(-34)\ Js`

m is the mass of electron having value
`9.11* 10^(-31)\ kg`

v is the speed of electron.

Given that v = c / 100

Where, c is the speed of light having value
`3* 10^8\ m/s`

Thus, v =
`\frac {3* 10^8}{100}\ m/s=3* 10^6\ m/s`

Applying in the equation as:

`\lambda=\frac {h}{m* v}`

`\lambda=\frac {6.626* 10^(-34)}{9.11* 10^(-31)* 3* 10^6}\ m`

`\lambda=(10^(-34)* \:6.626)/(10^(-25)* \:27.33)\ m`

`\lambda=(6.626)/(10^9* \:27.33)\ m`

`\lambda=2.42* 10^(-10)\ m`

Also,

`\\u=\frac {c}{\lambda}`

So,

`\\u=\frac {3* 10^8}{2.42* 10^(-10)}\ s^(-1)`

`\\u=(10^(18)* \:3)/(2.42)\ s^(-1)`

`\\u=1.24* 10^(18)\ s^(-1)`

Answer 2

DeBroglie wavelength is
`2.42* 10^(-10)\ m`

Frequency is
`1.24* 10^(16)\ Hz`

Step-by-step explanation:

Mass of electron = 9.10938356 × 10⁻³¹ kg

Planck's constant = h = 6.626 × 10⁻³⁴ m²kg/s

Speed of light = c = 3×10⁸ m/s

`\lambda=(h)/(p)=(h)/(mv)\\\Rightarrow \lambda=(6.626* 10^(-34))/(9.10938356* 10^(-31)* (3* 10^8)/(100))\\\Rightarrow \lambda=2.42* 10^(-10)\ m`

DeBroglie wavelength is
`2.42* 10^(-10)\ m`

`v=f\lambda\\\Rightarrow f=(v)/(\lambda)\\\Rightarrow ((3* 10^8)/(100))/(2.42* 10^(-10))\\\Rightarrow f=1.24* 10^(16)\ Hz`

Frequency is
`1.24* 10^(16)\ Hz`