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Vikki [24]
2 years ago
12

Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

ted?
Physics
1 answer:
marissa [1.9K]2 years ago
5 0

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})

where

E_0 = 13.6 eV is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV

Converting into Joules,

E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J

And now we can find the wavelength of the emitted photon by using the equation

E=\frac{hc}{\lambda}

where h is the Planck constant and c is the speed of light. Solving for \lambda,

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm

And this wavelength corresponds to red light.

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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
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Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

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Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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2 years ago
The average speed between earth and the sun is 1.50 x10^8 km. Calculate the average speed of the Earth in its orbit in kilometer
cluponka [151]

Answer:

The average speed of the earth in its orbit is 29.86km/s

Explanation:

The average distance between the Earth and the Sun is 1.50x10^{8} km.

The average speed of the earth in its orbit can be found by the next equation :

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In this case, the orbit of the Earth can be considered as a circle

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Then, equation 1 can be used:

v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}

v = 29.86km/s

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