Question

If a proton (1.673 X 10^-27 kg) is observed to have a wavelength of 1.65 X 10^-4 mm, what is its speed in meters per second?

Answer 1

moving particles like protons.electrons etc have kinetic energy and wavelength
.So wee need to combine both properties.De broglie equation explains the relation
between matter & wave properties

Use kinetic energy equation and debroglie equation
E= [1/2]mv^2 is the kinetic energy of proton
rearranging above equation
2E = mv^2

2E*m = m^2V^2

2Em = [mv]^2

mv = sqroot[2Em]
where m= mass,v= velocity
we have momentum p = mv
Thus p = sqroot[2Em]

Now De Broglie equation
wavelength lambda= h/p
plugin above equation
we have

wavelength lambda = h/sqroot[2Em]
now using given numbers find out energy first

given
wavelength = 1.65 X 10-4 mm =1.65 X 10-7 meters
mass m=(1.673 X 10-27 kg)

h=6.626 x 10¯34 J s

1.65 X 10-7 meters =6.626 x 10¯34 J s/sqroot[2E*(1.673 X 10-27 kg)]
squaring on both side

2.72 x10^-14 = 43.90 x10^-68/2E(1.673 X 10-27 k)

2E(1.673 X 10-27 )*2.72 x10^-14 =43.90 x10^-68
E*9.1 x10^-41 =43.90 x10^-68
E =43.90 x10^-68/9.1x10^-41 =4.82 x10^-27 J

Now using kinetic energy equation we can get velocity

E = 1/2{mv^2]

4.82 x10^-27 J= 1/2[(1.673 X 10-27 kg) v^2
v^2 =1/2[(1.673 X 10-27 kg) *4.82 x10^-27 J

V^2 = 0.062 x10^54 =6.20 x10^52

V = 2.49 x10^26 m/s
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