The de Broglie expression gives us:
sf(lambda_1=h/(mv_1))" "color(red)((1))
When the electron of mass m and charge e is accelerated through a potential difference of sf(V_1) volts we get:
sf(eV_1=1/2mv_1^2)" "color(red)((2))
Where sf(v_1) is the velocity of the electron.
The accelerating voltage is now increased by a factor of 4.
From sf(color(red)((2)) this means that the kinetic energy of the electron must also be increased by a factor of 4.
Since the kinetic energy of the electron depends on sf(v_1^2) this means that the velocity of the electron has increased by a factor of 2. (2 x 2 = 4).
From sf(color(red)((1))) you can see that doubling the velocity must, therefore, decrease the wavelength sf(lambda) by a factor of 2.
A more formal proof is given below:
From sf(color(red)((2))rArr)
:.sf(v_1^2=(2eV_1)/(m))
:.sf(v_1=sqrt((2eV_1)/(m)))
Subs. into sf(color(red)((1))rArr)
sf(lambda_1=(h)/(m.sqrt((2eV_1)/(m))
The voltage is now increased by a factor of 4:
sf(V_2=4V_1)
:.sf(lambda_2=(h)/(m.sqrt(2e.4V_1)/(m)))
sf(lambda_1/lambda_2=(cancel(h))/(cancel(m).sqrt((2eV_1)/(m)))xxcancel(m).sqrt((8eV_1)/(m))/cancel(h))
sf(lambda_1/lambda_2=sqrt((cancel(8)cancel(e)cancel(V_1))/(cancel(m))xx(cancel(m))/(cancel(2)cancel(e)cancel(V_1))))
sf(lambda_1/lambda_2=sqrt(4)=2)
:.sf(lambda_2=lambda_1/2)
This seems a reasonable result, as increasing the voltage should increase the kinetic energy and hence, the velocity of the electron.
The de Broglie expression tells us that this must decrease the wavelength.
