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1 year ago

What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?

1 answer:

7 0

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

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When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink

Dovator [93]

Answer:

y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

Therefore the diffraction equation for slits with m = 1 remains

a sin θ = λ

in general these experiments occur for oblique angles so

sin θ = θ

θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

θ = s / R

as the angle is small the arc approaches the distance s = y

y / R = 1.22 λ / s

y = 1.22 λ R / a

let's calculate

y = 1.22 500 10⁻⁹ 0.42 / 0.032

y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

y <8 10⁻⁶ m

4 0

3 years ago

Match each description to its written source.

madreJ [45]

1=6, 2=8
I hope this helped

3 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

2 years ago

PLEASE HELP!<br>I need help with this

melisa1 [442]

At time = .003 hr

90* (3*10^-3) = .27km

.005 hr

Car A = 150 * (.005) = .75km

Car B = (90 * (.005) = .45 km

Car A = 150 * (0.008) = 1.2 km

Car B = 90 * (0.008) = 0.72 km

Car A is ahead

3 0

2 years ago

How do our daily activities expose us to radiation?

dusya [7]

Explanation:

when we go outside, heat up food in the microwave, etc. it exposes us to radiation. along with things like radiators. the sun has lots of radiation and heat, so going outside exposes us. the microwave has radiation that heats up our food. the radiation from a microwave is not healthy for us, and is slowly poisoning us each day we use it. but as long as we do those things, we are exposed.

hope this helps

8 0

3 years ago