A particle (q = +5.0 µC) is released from rest when it is 2.0 m from a charged particle which is held at

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

2 years ago

If 30ml of a fluid has a mass of 63g what is it’s density

murzikaleks [220]

Answer:

The answer to your question is 2.1 g/ml

Explanation:

Data

volume = 30 ml

mass = 63 g

density = ?

Process

Density is defined as the mass per unit volume. The units of density are g/ml or kg/m³.

Formula

Density = mass / volume

Substitution

Density = 63 / 30

Result

Density = 2.1 g/ml

7 0

2 years ago

An iron bal of mass 20kg is rolling on a flat surface. On applying force, the velocity change from 17ms to 27m/s in 5s. Calculat

pentagon [3]

Answer:

40 N

Explanation:

We first need to calculate the acceleration of the tron ball.

Since acceleration, a = (v - u)/t where u = initial velocity of iron ball = 17m/s, v = final velocity of iron ball = 27m/s and t = time taken for the change in velocity = 5 s.

So, a = (v - u)/t

= (27 m/s - 17 m/s)/5 s

= 10 m/s ÷ 5 s

= 2 m/s²

We know force on iron ball, F = ma where m = mass of iron ball = 20 kg and a = acceleration = 2 m/s²

So, F = ma

= 20 kg × 2 m/s²

= 40 kgm/s²

= 40 N

So, the magnitude of the force on the iron ball is 40 N.

4 0

2 years ago

A negative charge of -3.0 x 10^-5 C is placed 0.35 m from a positive charge of 9.0 x 10^-5 C. What is the force between the two

Ilia_Sergeevich [38]

Answer:

the force between the two charges is 198.37 N

Explanation:

Given;

charge of the first particle, Q₁ = -3 x 10⁻⁵ C

charge of the second particle, Q₂ = 9.0 x 10⁻⁵ C

distance between the two charges, r = 0.35 m

The force between the two charges is calculated from Coulomb's law;

$F = \frac{kQ_1Q_2}{r^2}$

where k is Coulomb's constant = 9 x 10⁹ Nm²/C²

$F = \frac{(9\times 10^9)(3.0\times 10^{-5})(9\times10^{-5} )}{0.35^2} \\\\F = 198.37 \ N$

Therefore, the force between the two charges is 198.37 N

8 0

1 year ago

Is it true that when travelling at speed of light , time stops?

kolbaska11 [484]

Answer: Yes, the time stops when traveling at the speed of light.

Explanation:

7 0

2 years ago