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Sidana [21]
3 years ago
5

The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

ed. If the wavelength of the emitted gamma ray is 3.44×10−12 m , what is the decrease in mass, ΔmXe, of the xenon nucleus as a result of this decay?
Physics
1 answer:
nevsk [136]3 years ago
6 0

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

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Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

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- The time period for 1 revolution T = 24 hrs

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What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

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                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
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Answer:

r=5.278\times 10^{-4}\ m

Explanation:

Given that:

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KE= \frac{1}{2} m.v^2

1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2

v^2=4.2198\times 10^{11}

v=6.496\times 10^6\ m.s^{-1}

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r=\frac{m.v}{q.B}

r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}

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