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aev [14]
2 years ago
7

A proton (1.6726 ? 10-27 kg) and a neutron (1.6749 ? 10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or "

heavy hydrogen." In this process, a gamma ray (high-energy photon) is emitted, and its energy is measured to be 2.2 MeV (2.2 ? 106 eV).
(a) Keeping all five significant figures, what is the mass of the deuteron? Assume that you can neglect the small kinetic energy of the recoiling deuteron.

(b) Momentum must be conserved, so the deuteron must recoil with momentum equal and opposite to the momentum of the gamma ray. Calculate approximately the kinetic energy of the recoiling deuteron and show that it is indeed small compared to the energy of the gamma ray.
Physics
1 answer:
Alexus [3.1K]2 years ago
5 0

Answer:

Explanation:

The mass of the deuteron = mass of the proton + mass of the neutron + mass equivalent of the energy of 2.2 Mev evolved.

I amu = 931 Mev

2.2 Mev = 2.2 / 931 amu

= ( 2.2 / 931 )x 1.6726 x 10⁻²⁷

= .00395 x 10⁻²⁷

The mass of the deuteron  =( 1.6726 + 1.6749 +  .00395)x 10⁻²⁷ kg

= 3.35145 x 10⁻²⁷ kg

b ) Momentum of gamma ray

= h / λ ( h is plank's constant and λ is wavelength of gamma ray )

= hυ / υλ       (  υ is frequency of gamma ray )

= E / c  ( E is energy of photon and c is velocity o light )

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹  J / 3 x 10⁸

= 1.173 x 10⁻²¹ Kg m /s

This will be the momentum of deuteron also

Kinetic energy

= p² / 2m ( p is momentum and m is mass of deuteron )

= ( 1.173 x 10⁻²¹ )² / ( 2 x 3.35145 x 10⁻²⁷)

= 1.376 x ⁻¹⁵ J

Energy of gamma ray

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹  J

= 3.52 x 10⁻¹³ J

So kinetic energy of deuteron is smaller than energy of gamma ray photon .

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A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the
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Answer:

a)R = 171μΩ

b)E = 1.7 *10^{-4} V/m

c)R_{2} = 1.16 *10^{-4}Ω

here * stand for multiplication

Explanation:

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radius of cylinder  =  1.1 cm

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(here * stand for multiplication )

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b = 2.78* 10^{-8} Ωm

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E = \frac{[a+b(L/2)^2]I }{\pi  r^2}

by substitute the values

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E = 1.7 *10^{-4} V/m

(here * stand for multiplication )

c ).

75 cm means length will be half

 that is   x =  L/2

integrate  the second equation with upper limit  L/2  

Let resistance is R_{1}

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R_{1}  =  \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}

substitute the value of a , b and L we get

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R_{2} = 1.16 *10^{-4}Ω

(here * stand for multiplication )

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