Question

Li2+ and a proton are accelerated by the same potential, their de-Broglie wavelength have the ratio (assume mass of proton = mass of neutron):

​

Answer 1

Answer:

m= mass

q= charge

V= potential

λ=h/mv

λ

proton

=h/mv

The momentum acquired by the proton = (2mqV)

1/2

Thus λ

proton

=h/(2mqV)

1/2

For λ

Be

+3

=h/(2×9m×3q×V)

1/2

Then λ

Be

+3

/λ

proton

=1/3(3)

1/2