What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV

(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)

λ = 0.173 x 10^ (-9) m = 0.173 nm

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s) / (0.173 x 10^ (-9) m)

f = 2.42 x 10^16 Hz

(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)

λ = 2.875 x 10^ (-12) m = 2.875 pm

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s) / (2.875 x 10^ (-12) m)

f = 4.8 x 10^16 Hz