Question

An alpha particle is accelerated through a potential difference of v volts from rest.the de broglie wavelength associated with it

Answer 1

Answer:

Explanation:

Let the wavelength be X

Now we know that

K.E of an alpha particle is equal to the product of charge and potential diff

Thus 1/2mv^2 = q*v

X=h/mv

X=h/✓(m^2*v^2)

X=h/✓(1/2*m*v^2*(2m))

X=h/✓(K.E*(2m))

X=h/✓(2m*q*v)

Now putting the values of m and q

m=4*1.6*10^-27

q=2*1.6*10^-19

Thus X=(0.103*10^-10)/✓v