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3 years ago

Draw the vector C⃗ =A⃗ +2B⃗ .

Physics

1 answer:

spin [16.1K]3 years ago

4 0

Answer:

A) C is 2 units towards right

B) C is 9 units towards right and 3 units towards up

C) C is 4 units towards right and 1 unit towards down

Explanation:

Drawings are given in the attachment.

A)Since,

A is 4 units towards right,

B is 1 unit towards left,

Then,

C is 2 units towards right.

B) Since,

A is 4 units towards right and 2 units towards up,

B is 1 unit towards left,

Then,

C is 9 units towards right and 3 units towards up.

C) Since,

A is 4 units towards right and 2 units towards up,

B is 1 unit towards right, and 1 unit towards down

Then,

C is 4 units towards right and 1 unit towards down.

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Answer:

2,8

Explanation:

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3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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Answer:

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Explanation:

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