All categories

4 years ago

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

Engineering

1 answer:

DaniilM [7]4 years ago

6 0

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)

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

You might be interested in

A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The

wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

$q=\frac{Ak}{L}dT$

$\frac{q}{A}=\frac{k}{L}dT$

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

${k_{fi}= 0.03 W/m.K$

${L_{fi}= 5 cm = 0.05 m$

${T_{fi}= 25 \°C$

${k_{cb} = 0.5 W/m.K$

${L_{cb}= 20 cm = 0.20 m$

${T_{cb}= 0 \°C$

${T_{m}= ? \°C =$ temperature at the interface

Solving for ${T_{m}$ will give the temperature at the interface:

$\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})$

$\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})$

$15 -0.6T_{m}=2.5T_{m}$

$3.1T_{m}=15$

$T_{m}=4.8$

3 0

3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

3 years ago

Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M

Naddika [18.5K]

Answer:

a. 318.2k

b. 45.2kj

Explanation:

Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

7 0

3 years ago

Which is the better measure of computer system performance—a benchmark, such as SPECINT; or a processor speed measure, such as G

Vesna [10]

Answer:

A benchmark

Explanation:

Most times a benchmark serves as the better measure when assessing a computer's performance, this is because CPU speeds can only evaluate an aspect of a computer's performance whereas a benchmark offers the advantage of measuring all the aspects of a computer's performance for a specific type of computing problem.

5 0

3 years ago

The increasing interconnections of peoples and countries around the world is known as ________.

Irina18 [472]

I think it’s rationalization.
Hope this helps

4 0

3 years ago

Read 2 more answers