A proton (p = +, mp = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron ( = +, m = 2.0 u) and an alpha particle ( = +2, m = 4.0 u) are accelerated from rest through the same potential difference , and then enter the same region of uniform magnetic field ⃗⃗ , moving perpendicularly to the direction of the magnetic field.
a. What is the ratio of the proton’s kinetic energy Kp to the alpha particle’s kinetic energy K
b. What is the ratio of the deuteron’s kinetic energy K to the alpha particle’s kinetic energy K
c. If the radius of the proton’s circular orbit p = 10 cm, what is the radius of the deuteron’s orbit
d. What is the radius of the alpha particle’s orbit

a. 1/2 b. 1/2 c, 20 cm d. 40 cm

Explanation:

Here is the complete question

A proton ( = +, = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron ( = +, = 2.0 u) and an alpha particle ( = +2, = 4.0 u) are accelerated from rest through the same potential difference , and then enter the same region of uniform magnetic field ⃗⃗ , moving perpendicularly to the direction of the magnetic field.

A) What is the ratio of the proton’s kinetic energy to the alpha particle’s kinetic energy ?

B) What is the ratio of the deuteron’s kinetic energy to the alpha particle’s kinetic energy ?

C) If the radius of the proton’s circular orbit = 10 cm, what is the radius of the deuteron’s orbit ?

D) What is the radius of the alpha particle’s orbit ?

Solution

a. For both particles, kinetic energy = electric potential energy

For proton K.E= K₁ = 1/2m₁v₁² = +eV , for alpha particle K.E = K₂ = 1/2m₂v₂²= +2eV

where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the proton and alpha particle. So, the ratio of their kinetic energies is

1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV

m₁v₁²/m₂v₂² = 1/2.

So the ratio K₁/K₂ = 1/2

b. For both particles, kinetic energy = electric potential energy

For deuteron  K₁ = 1/2m₁v₁² = +eV , for alpha particle K₂ = 1/2m₂v₂²= +2eV

where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the deuteron and alpha particle. So, the ratio of their kinetic energies is

1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV

m₁v₁²/m₂v₂² = 1/2.

So the ratio K₁/K₂ = 1/2

c. The radius of the proton's circular is gotten from the centripetal force which equal the magnetic force. So,

mv²/r = Bev

r₁ = mv/Be

Since mass of deuteron m₂ equals twice mass of proton m₁, m₂ = 2m₁

So, radius of deuteron's circular orbit equals

r₂ = m₂v/Be = 2m₁v/Be = 2r₁ = 2 × 10 cm = 20 cm

d. The radius of the alpha particle is given by r₃ = m₃v/Be. Since mass of alpha particle equal four times mass of proton, m₃ = 4m₁.

So, radius of alpha particle's circular orbit equals

r₃ = m₃v/Be = 4m₁v/Be = 4r₁ = 4 × 10 cm = 40 cm

ill do it for 100, also this isnt 80 its 40

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27