Question

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV, (c) 100 kV.

Answer 1

(a)
`\lambda=1.227\ A`

(b)
`\lambda=0.388\ A`

(c)
`\lambda=0.038\ A`

Step-by-step explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

`\lambda=(h)/(√(2meV) )`

Where

m and e are the mass of and charge on an electron

On solving,

`\lambda=(12.27)/(√(V) )\ A`

V = 100 V

`\lambda=(12.27)/(√(100) )\ A`

`\lambda=1.227\ A`

(b) V = 1 kV = 1000 V

`\lambda=(12.27)/(√(V) )\ A`

`\lambda=(12.27)/(√(1000) )\ A`

`\lambda=0.388\ A`

(c) If
`V=100\ kV=10^5\ V`

`\lambda=(12.27)/(√(10^5) )\ A`

`\lambda=0.038\ A`

Hence, this is the required solution.