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mars1129 [50]
3 years ago
14

If a nucleus decays by successive b, a, a emissions, its mass number will Group of answer choices decrease by seven. decrease by

two. decrease by four. decrease by eight. increase by four.
Physics
1 answer:
nirvana33 [79]3 years ago
7 0

Answer:

The mass number will decrease by eight (8).

Explanation:

Given;

successive beta (b), alpha (a), alpha (a) emissions.

Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 . (^{0}_{-1}\beta )

Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2. (^4_2\alpha)

For the given question, a successive beta (b), alpha (a),  and alpha (a) emissions = (0) + (-4) + (-4) = -8

Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).

<u>To learn more, please visit:</u> brainly.com/question/23276969

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C.The traditional flow of current is positive to negative and the flow of electrons is the opposite

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The granulation pattern that astronomers have observed on the surface of the Sun tells us that: a. the Sun is a lot cooler on th
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c. hot material must be rising from the Sun's hotter interior

Explanation:

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Calculate the kinetic energy of a moving 4 kg object traveling at a velocity of 3 m/a
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3 years ago
Calculate the energy per photon and the energy per mole of photons for radiation of wavelength (a) 200 nm (ultraviolet), (b) 150
VMariaS [17]

Answer: a. E =9.9*EXP(-19)J

1 mole E= 596178J

b. E= 1.32*EXP(-15)J, 1 mole E=795MegaJ

c. E= 1.98*EXP(-23)J

1 mole E = 11.9J

Explanation: The Energy of a photon E, the wavelength are related by

E= h*c/wavelength

h is the Planck's constant 6.6*EXP(-34)J.s

c is speed of light 3*EXP(8)m/s

h*c=1.98*EXP(-25)

Now let's solve

a. E = h*c/wavelength

= h*c/(200*EXP(-9)m

=9.9*EXP(-19)J

1 mole of a photon contian 6.022*EXP(23)photons by advogadro

Now to get the energy of 1 mole of the photon we have

9.9*EXP(-19)*6.023*EXP(23)

=596178J

b. E=h*c/150*EXP(-12)m

=1.32*EXP(-15)J

1 mole will have

1.32*EXP(-15)*6.022*EXP(23)J

=795*EXP(6)J

c. E= h*c/1*EXP(-2)m

=1.98*EXP(-23)J

1 mole of the photon will have

1.98*EXP(-23)J *6.022*EXP(23)

= 11.9J.

You will notice that the longer the wavelength of the photon the lesser the Energy it as.

NOTE: EXP represent 10^

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