Answer:
(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz
(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz
Explanation:
(a)
First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,
K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)
(1/2)mv² = 8 x 10^(-18) J
Mass of electron = m = 9.1 x 10^(-31) kg
Therefore,
v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)
v = √1.75 x 10^13
v = 4.2 x 10^6 m/s
Now, the de Broglie's wavelength is given as:
λ = h/mv
where,
h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s
Therefore,
λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)
<u>λ = 0.173 x 10^(-9) m = 0.173 nm</u>
The frequency is given as:
Frequency = f = v/λ
f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)
<u>f = 2.42 x 10^16 Hz</u>
(b)
First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,
K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)
(1/2)mv² = 1.6 x 10^(-17) J
Mass of proton = m = 1.67 x 10^(-27) kg
Therefore,
v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)
v = √1.916 x 10^10
v = 1.38 x 10^5 m/s
Now, the de Broglie's wavelength is given as:
λ = h/mv
where,
h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s
Therefore,
λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)
<u>λ = 2.875 x 10^(-12) m = 2.875 pm</u>
The frequency is given as:
Frequency = f = v/λ
f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)
<u>f = 4.8 x 10^16 Hz</u>
