PHYSICS COLLEGE

When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum

Answers

Answer 1

Answer:

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

$E = hf = h(c)/(\lambda) \n\nwhere;\n\nh \ is \ Planck's \ constant = 6.626 * 10^(-34) \ Js\n\nc \ is \ speed \ of \ light = 3 * 10^(8) \ m/s\n\nE = ((6.626* 10^(-34))* (3* 10^8))/(248* 10^(-9)) \n\nE = 8.02 * 10^(-19) \ J$

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)

Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

$\phi = hf = (hc)/(\lambda_(max)) \n\n\lambda_(max) = (hc)/(\phi) \n\n\lambda_(max) = ((6.626* 10^(-34)) * (3 * 10^8) )/(6.546 * 10^(-19)) \n\n\lambda_(max) = 3.037 * 10^(-7) m\n\n\lambda_(max) = 303.7 \ nm$

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An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:Find the radius and period of the orbit.
A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?
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A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the tires rotating? Part (b) What is the centripetal acceleration at the edge of the tire in m/s^2?

Answers

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =(72)/(0.445)=161.8rad/s$

Frequency

$f=(2\pi)/(\omega)=(2* \pi)/(161.8)=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

$a=(v^2)/(r)$

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=(72^2)/(0.445)=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

2H is a loosely bound isotope of hydrogen, called deuterium or heavy hydrogen. It is stable but relatively rare — it form only 0.015% of natural hydrogen. Note that deuterium has Z = N, which should tend to make it more tightly bound, but both are odd numbers.Required:
Calculate BE/A, the binding energy per nucleon, for 2H in megaelecton volts per nucleon

Answers

Answer:

0.88 MeV/nucleon

Explanation:

The binding energy (B) per nucleon of deuterium can be calculated using the following equation:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u$

Where:

Z: is the number of protons = 1

N: is the number of neutrons = 1

$m_(p)$ : is the proton's mass = 1.00730 u

$m_(n)$ : is the neutron's mass = 1.00869 u

M: is the nucleu's mass = 2.01410

A = Z + N = 1 + 1 = 2

Now, the binding energy per nucleon for ²H is:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u = (1*1.00730 + 1*1.00869 - 2.01410)/(2)*931.49 MeV/u = 9.45 \cdot 10^(-4) u*931.49 MeV/u = 0.88 MeV/nucleon$

Therefore, the binding energy per nucleon for ²H is 0.88 MeV/nucleon.

I hope it helps you!

Final answer:

The binding energy per nucleon for 2H (deuterium) is 1.1125 MeV per nucleon.

Explanation:

The binding energy per nucleon, or BE/A, can be calculated by dividing the total binding energy of the nucleus by the number of nucleons. To calculate the BE/A for 2H (deuterium), we need to know the total binding energy and the number of nucleons in deuterium. The total binding energy of deuterium is approximately 2.225 MeV (megaelectron volts) and the number of nucleons is 2. Therefore, the BE/A for 2H is 2.225 MeV / 2 = 1.1125 MeV per nucleon.

Learn more about Binding energy per nucleon here:

brainly.com/question/10095561

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A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/s

E. None of the option is correct

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer:

$E = -9.5* 10^4~N/C$

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

$\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)$

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

$E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C$

How do impacts by comets and asteroids influence Earth’s geology, its atmosphere, and the evolution of life?

Answers

Answer:

Explanation:

A comet is a celestial body made up of ice and dust and assumed to have a tail.

As per considering ancient history, they are even termed as death-dealers and have doomed many planets.

If we talk about Earth, comets and asteroids have played a massive role in changing Earth's geology, atmosphere and evolution. Its believed that dinosaurs population is wiped out from Earth because of comets that falls during that time. Many new diseases are even brought by them.

Large comets can result into global environmental damage and can even lead to mass extension.The dust from the impact and the heat creates many harmful oxides resulting into acid rain and can kill thousand of organism.

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