Ask question

Ask question

All categories

2 years ago

10

Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r

adiation?

2 answers:

Helga [31]2 years ago

7 0

We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

E = 1.578 x 10^ -24 J

kirza4 [7]2 years ago

3 0

The energy of exactly one photon of this microwave radiation is about 1.58 × 10⁻²⁴ J

$\texttt{ }$

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

$\large {\boxed {E = h \times f}}$

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

$\texttt{ }$

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

c = 3 × 10⁸ m/s

h = 6.63 × 10⁻³⁴ Js

λ = 12.6 cm = 0.126 m

<u>Unknown:</u>

E = ?

<u>Solution:</u>

$E = h \times f$

$E = h \times \frac{c}{\lambda}$

$E = (6.63 \times 10^{-34}) \times \frac{3 \times 10^8}{0.126}$

$E = 1.58 \times 10^{-24} \texttt{ J}$

$\texttt{ }$

<h3>Learn more</h3>

Photoelectric Effect : brainly.com/question/1408276
Statements about the Photoelectric Effect : brainly.com/question/9260704
Rutherford model and Photoelecric Effect : brainly.com/question/1458544

$\texttt{ }$

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Quantum Physics

You might be interested in

A lever has a 36cm input arm and a 6cm output arm. find MA.

Nady [450]

6

Explanation:

The mechanical advantage is a factor that measures how input force increases using a machine.

A lever is a simple machine with the fulcrum at the center.

To calculate the mechanical advantage M.A of levers we use the expression below;

M. A = $\frac{F_{b} }{F_{a} }$ = $\frac{a}{b}$

$F_{a}$ = input force

$F_{b}$ = output force

a is the distance of the input force from the fulcrum

b is the distance of the output force from the fulcrum

Given

a = 36cm

b = 6cm

M.A = $\frac{36cm}{6cm}$ = 6

learn more:

Torque brainly.com/question/5352966

#learnwithBrainly

6 0

2 years ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu

Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0

1 year ago

The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and

mina [271]

Answer:

$\frac{dQ}{dt}= 4312 W$

Explanation:

As we know that base of the slab is given as

$A = 11 \times 8$

$A = 88 m^2$

now we know that rate of heat transfer is given as

$\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)$

here we know that

$k = 1.4 W/m k$

Also we have

$x =0.20$

$\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)$

$\frac{dQ}{dt}= 4312 W$

7 0

2 years ago

A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul

Rashid [163]

The Young modulus is given by:
$E= \frac{F /A}{\Delta L / L_0}$
where
F is the force applied
$L_0$ is the initial length of the wire
$A$ is the cross-sectional area of the wire
$\Delta L$ is the stretch of the wire

The wire in the problem stretches by $0.5%$ of its length, this means
$\frac{\Delta L}{L_0} = 0.005$

We can also calculate the area of the wire; its radius is in fact half the diameter:
$r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m$
and so the area is
$A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2$

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

$E= \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2$

3 0

1 year ago