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melamori03 [73]
3 years ago
15

When an atom releases gamma radiation____________.

Physics
1 answer:
wel3 years ago
3 0

Answer:

d. the atomic number remains the same.

Explanation:

First at all it's important to know how to read nuclear information

_{Z}^{A}X

X is the atomic symbol, A the mass number and Z the atomic number of the element.

Gamma rays emitted on gamma decay are characterized as _{0}^{0}\gamma

If we write the nuclear equation for the decay, we have that:

_{Z}^{A}X\,\rightarrow{}_{0}^{0}\gamma\,+\,{}_{Z}^{A}Y

The sum of the mass numbers and atomic numbers on the right side has to be equal to the left side numbers of the equation, that means the mass number and the atomic number remains the same for the resulting atom to preserve the equality.

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u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled

Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
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Answer: 1.95 x 10¹⁴ m/s²

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Willingness to take turns is one way we can express our attitudes in
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Write the importance of international bureau of weights and measurement.​
Taya2010 [7]

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8 0
3 years ago
A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

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We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

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2 years ago
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