Question

What is the wavelength in picometers for a neutron moving at 753 ?

Answer 1

`\lambda=525.37\ pm`

Step-by-step explanation:

Given,

Velocity of the neutron = 753 m/s

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 neutron having value
`1.6749* 10^(-27)\ kg`

v is the speed of neutron = 753 m/s

Applying in the equation as:

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

`\lambda=(6.626* 10^(-34))/(1.6749* 10^(-27)* 753)\ m`

`\lambda==(10^(-34)* \:6.626)/(10^(-27)* \:1261.1997)\ m`

`\lambda==(6.626)/(12611997000)\ m`

`\lambda=5.2537* 10^(-10)\ m`

1 m = 10¹² pm

So,

`\lambda=525.37\ pm`