58.6k views
0 votes
3. Find the de Broglie wavelength of the following particles: (i) An electron in a semiconductor having average thermal velocity at T = 300 K

User Nck
by
2.9k points

2 Answers

2 votes

Final answer:

The de Broglie wavelength of an electron can be calculated using de Broglie's equation, λ = h/p. To find the wavelength for an electron in a semiconductor at 300 K, one needs to know the electron's velocity, which is related to its kinetic energy at that temperature. Precise calculation would require details about the effective electron mass in the semiconductor.

Step-by-step explanation:

The de Broglie wavelength λ of a particle can be determined using de Broglie's equation, λ = h/p, where 'h' is Planck's constant (6.626 × 10-34 m2 kg/s) and 'p' is the momentum of the particle. The momentum of an electron moving at thermal velocities in a semiconductor at room temperature (T = 300 K) can be approximated using the formula p = mv, where 'm' is the mass of an electron (9.11 × 10-31 kg) and 'v' is the velocity. The average thermal velocity can be obtained from the kinetic theory of gases, which states that the average kinetic energy (KE) of a particle is ⅓kT, where 'k' is the Boltzmann constant (1.38 × 10-23 J/K) and 'T' is the temperature in kelvins. Therefore, v can be calculated as √(3kT/m). Substituting this velocity into the momentum formula and then into de Broglie's equation yields the de Broglie wavelength of the electron.

However, a precise value for the de Broglie wavelength of an electron in a semiconductor at T = 300 K would require additional information not provided in the question, such as the effective mass of the electron in the semiconductor, which can differ from the free electron mass due to the crystal structure's influence.

User Iqra
by
3.6k points
3 votes

Answer:

The wavelength of the electron is
(4.4097*10^(-9))/(√(a))\ m

Step-by-step explanation:

Given that,

Temperature = 300 K

We know that,

The energy of free electron is


E=((\hbar)^2k^2)/(2m)


k=(√(2mE))/(\hbar)

Where, k = wave number

The momentum of the electron is


p=\hbar k

Th effective mass is


m=am_(0)

We need to calculate the wavelength of the electron

Using formula of wave number


k=(2\pi)/(\lambda)


\lambda=(2\pi)/(k)

Put the value of k


\lambda=(2\pi)/((√(2mE))/(\hbar))


\lambda=\frac{h}{\sqrt{2am_(0)E}}

We know that,

Thermal energy of electron


E=3kT

The de Broglie wavelength of the electron is


\lambda=\frac{h}{\sqrt{2am_(0)*3kT}}

Put the value into the formula


\lambda=\frac{6.63*10^(-34)}{\sqrt{2*9.1*10^(-31)*3*300*1.380*10^(-23)* a}}


\lambda=(4.4097*10^(-9))/(√(a))\ m

Hence, The wavelength of the electron is
(4.4097*10^(-9))/(√(a))\ m

User Drenda
by
3.4k points