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RUDIKE [14]
10 months ago
11

light of wavelength 3.0 × 10−7 m shines on the metals lithium, iron, and mercury, which have work functions of 2.3 ev, 3.9 ev, a

nd 4.5 ev, respectively. which of these metals will exhibit the photoelectric effect? the speed of light is 3 × 108 m/s and planck’s constant is 6.63 × 10−34 j · s.
Physics
1 answer:
Greeley [361]10 months ago
8 0

None of these will show photoelectric effect as for photoelectric effect energy must be greater than work function.

a stream of particles whose energy follow the Planck formula for their frequencies. The photons and atoms collide when that beam strikes a metal. The photoelectric effect is created by a collision when a photon's frequency is high enough to remove an electron.

Here we have;

Light of wavelength = 3× 10∧-7

Speed of light= 3×10∧8

plank's constant= 6.63×10∧-34

Therefore, E=hc/lambda=6.63×10∧-19=6.63ev

as you can  see none of wave function is greater than energy so no one will show photoelectric effect. The frequency of the incident radiation and the surface material are the two elements controlling the maximum kinetic energy of photoelectrons.

Learn more about photoelectric effect here; brainly.com/question/9260704?

#SPJ4

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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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