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

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Answer:


\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

User Jordan Breton
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