PHYSICS HIGH SCHOOL

What is the energy of a photon whose frequency is 6.0 x 10^20?(h=6.63 x10^-34 j•s; 1 eV=1.06x10^-19 J)

A. 1.6 MeV

B. 2.5 MeV

C. 3.3 MeV

D. 4.8 MeV

Answers

Answer 1

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

The correct answer is not in the options, but the closest to it is option C.

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B.no change occurs in the number of wavelengths that pass a given point per second

C.a greater number of wavelengths pass a given point per second

D.the number of wavelengths that pass a given point per second doubles

Answers

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The charge on a gamma ray is ____.
a. 2
b. 1
c. 0
d. 2

Answers

So we want to know what charge do gamma particles have. Gamma particles are high energy electromagnetic radiation or in other words, gammy rays are high energy photons. Photons have no electrical charge so the charge of gamma rays is equal to zero. The correct answer is c. 0

Answer:

1) An unstable nucleus

c. emits energy when it decays

2)The charge on gamma ray is

c. 0

3) what particle is emitted in alpha radiation?

c. helium nucleus

4) The least penetrating form of radiation is

c. Alpha radiation

5) What is the change in atomic number when an atom emits a BETA particle?

D. increases by 1

- There you go loves :)

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a. True
b. False

Answers

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Answers

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

Here, we have been given the height through which the ball drops;

Height of drop = 14.5m - 1.9m = 12.6m

The right motion equation is;

V² = U² + 2gh

V is the final velocity

U is the initial velocity = 0

g is the acceleration due to gravity = 9.8m/s²

h is the height

Now insert the parameters and solve;

V² = 0² + 2 x 9.8 x 12.6

V² = 246.96

V = √246.96 = 15.7m/s

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Answers

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