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.
