PHYSICS COLLEGE

What is the wavelength of the electromagnetic radiation needed to eject electrons from a metal?

Answers

Answer 1

Answer:

Answer:

λ = hc/(eV + h $f_(0)$ )

Explanation:

Let the work function of the metal = ∅

the kinetic energy with which the electrons are ejected = E

the energy of the incident electromagnetic wave = hf

Then, we know that the kinetic energy of the emitted electron will be

E = hf - ∅

because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.

This means that the energy of the incident e-m wave can be written as

hf = E + ∅

also, we know that the kinetic energy of the emitted electron E = eV

and the work function ∅ = h $f_(0)$

we can they combine all equations to give

hf = eV + h $f_(0)$

we know that f = c/λ

substituting, we have

hc/λ = eV + h $f_(0)$

λ = hc/(eV + h $f_(0)$ ) This is the wavelength of the e-m radiation needed to eject electrons from a metal.

where

λ is the wavelength of the e-m radiation

h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s

c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s

e is the charge on an electron

V is the voltage potential on the electron

$f_(0)$ is the threshold frequency of the metal

Related Questions

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken
The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?
Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin film of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?
A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s 2 . What is the moment of inertia of the door about the hinges?

Answers

Answer:

Moment of inertia will be $I=2kgm^2$

Explanation:

We have given mass of the person m = 72 kg

Radius r = 0.8 m

Force is given F = 5 N

Angular acceleration $\alpha =2rad/sec^2$

Torque is given by $\tau =F* r=5* 0.8=4N-m$

We know that torque is also given by

$\tau =I\alpha$ , here I is moment of inertia and $\alpha$ is angular acceleration

So $4=I* 2$

$I=2kgm^2$

Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?

Answers

Explanation:

The relation between resistance and resistivity is given by :

$R=\rho (l)/(A)$

$\rho$ is resistivity of material

l is length of wire

A is area of cross section of wire

Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

Answers

Answer:

The no. of electrons is $7.59* 10^(21)$

Solution:

According to the question:

The rate at which the charge is delivered is given by:

$(dQ)/(dt) = - 0.75$

Now,

$\int_(0)^(Q)dQ = - 0.75\int_(0)^(27 min) dt$

$Q = -0.75t|_(0)^(27 min)$

$Q= -0.75* 27* 60 = - 1215 C$

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge = $1.6* 10^(- 19) C$

Thus

$n = (Q)/(e)$

$n= (1215)/(1.6* 10^(- 19))$

$n = 7.59* 10^(21)$

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

= $Tensile \ strength* Area$

On substituting the values, we get

= $(3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)$

= $(3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)$

= $21195 \ N$

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

Learn more about Tensile Strength here:

brainly.com/question/14293634

#SPJ3

What is science? Plz help I cant find a answer thats not really complicated and i have to fit it within 3 sentences.

Answers

Science is the study of our universe, and our own planet. It is the study of the biological, chemical and physical world we live in. It is a process of discovering by experimenting and looking for patterns.

Please show steps as to how to solve this problem
Thank you!

Answers

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1 * X2 =

62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y will be out of the page

r X F for F1 will be into the page so the torque must be negative

Other Questions

An artificial satellite circling the Earth completes each orbit in 144 minutes. (a) Find the altitude of the satellite.

Which is a characteristic of thermal energy transfer through convection

The tension in a string is 15 N, and its linear density is 0.85 kg/m. A wave on the string travels toward the -x direction; it has an amplitude of 3.6 cm and a frequency of 12 Hz. What are the: a. Speed b. Wavelength of the wave? c. Write down a mathematical expression for the wave, substituting numbers for the variables

A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.02500.0250-\text{s}s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. Determine the magnitude of the emf induced in the loop