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DanielleElmas [232]
3 years ago
14

A. Write 2 -3 sentences describing the mass number, charge, and penetrating ability of gamma rays.

Physics
2 answers:
Leya [2.2K]3 years ago
8 0

Answer:

a. Mass number is the sum of number of protons and neutrons present in the nucleus of an atom.

A = p + n

Charge is a unit of matter that tells how more or less the number of electrons are present as compared to number of protons in an atom. A positive charge means the atom lacks electrons.

Gamma rays are highly energetic electromagnetic rays. They have high penetrating power and are very difficult to shield owing to high energy.

b. The compete nuclear equation is:

^{210} _{89} Ac \rightarrow ^4 _2 He + ^{206} _{87}Fr

This is a alpha decay process because alpha particle which is Helium atom is being produced. Francium is being formed after alpha particle is emitted.

c. The complete nuclear reaction is:  

^{80} _{38} Sr \rightarrow ^0 _{+1}e + ^{80} _{37}Rb

This is a positron emission process where in Rubidium is being formed.

Evgen [1.6K]3 years ago
7 0

a) Their mass number is 0, the charge is 0, that is why only a thick wall of lead or concrete can protect from them.

b) Mass number is 206, charge is 87, it is an isotope of francium.

c) Mass number will remain 80, charge is 37, it is rubidium

You might be interested in
Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 &lt; x a. 283 eV <br> b. 339 eV <br> c.
denis23 [38]

This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113   eV  

d. 226 eV        

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

E_{nx,ny,nz = ( n_x^2 + n_y^2 + n_z^2 ) × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( n_x = n_y = n_z = 1 )

so

( n_x^2 + n_y^2 + n_z^2 ) ×  π²h"² / 2ml²

since h" = h/2π

( n_x^2 + n_y^2 + n_z^2 ) × π²h² / (2π)²2ml²

so we substitute

E_{111 = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

E_{111 = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]    

E_{111 = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

E_{111 = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

E_{111 = 3 × [ 37.645578 ]

E_{111 = 112.9 ≈ 113 eV

E_{211 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{211 = ( 1² + 1² + 2² ) × [ 37.645578 ]

E_{211 = 6 × [ 37.645578 ]

E_{211 = 225.87 ≈ 226 eV

E_{221 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{221 = ( 2² + 2² + 1² ) × [ 37.645578 ]

E_{211 = 9 × [ 37.645578 ]

E_{211 = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

8 0
2 years ago
The_____is the time it takes for a wave to complete one cycle​
Zepler [3.9K]

Answer:

I think the correct answer is period

3 0
3 years ago
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS
deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

v=\frac{2\pi r}{T} (1)

where

2\pi r is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: v=7,750 m/s. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m

So, we can re-arrange equation (1) to find the orbital period:

T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0
3 years ago
Read 2 more answers
A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit
Alecsey [184]

Answer:

  F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

          I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

        v = d / t

        t = d / v

Reduce SI system

          m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

          d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

         t = 50 10⁻³ / 600

         t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

        F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

       F = m (v-vo) / t

       F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

       F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

5 0
3 years ago
If transpiration didn't take place, would water be able to enter the roots of the plant? Explain your reasoning
s2008m [1.1K]

Answer:

If transpiration didn't take place water would still be able to enter the roots of a plant

Explanation:

transpiration is the process of water leaving from living organisms to the atmosphere, therefore, if transpiration didn't occur the water would not transpire to the atmosphere and would remain in the root but water absorption would not change because it is a biological need for the living organism as such

3 0
3 years ago
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