Question

A proton is accelerated to 225V. Its de-Broglie wavelength is:

a. 1.91 pm
b. 0.2 pm
c. 3 pm
d. 0.4 pm

Answer 1

The value is
`\lambda =  1.9109 *10^(-12) \  m`

Step-by-step explanation:

From the question we are told that

The potential of the proton is
`V = 225 \ V`

Generally the momentum of the particle is mathematically represented as

`p  =  √( 2 *  m  *  V  *  e )`

Here e is the charge on the proton with value

`e =  1.60 *10^(-19) \  C`

m is the mass of the proton with value
`m  =  1.67 *10^(-27) \  kg`

So

`p  =  \sqrt{ 2 * (1.67*10^(-27) ) *  225 *  1.6*10^(-19)}`

=>
`p  = 3.4676 *10^(-22) \  kg \cdot m/ s`

So the de-Broglie wavelength isis mathematically represented as

`\lambda  =  (h)/(p)`

Here h is the Planck's constant with value

`h = 6.626 *10^(-34) \  J\cdot s`

=>
`\lambda  =  (6.626 *10^(-34))/(3.4676 *10^(-22) )`

=>
`\lambda =  1.9109 *10^(-12) \  m`