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?
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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