All categories

2 years ago

Describe how electromagnetic radiation can ionise an atom. 2 marks

Physics

1 answer:

IRISSAK [1]2 years ago

8 0

Answer:

Ionizing radiation is radiation with enough energy so that during an interaction with an atom, it can remove tightly bound electrons from the orbit of an atom, causing the atom to become charged or ionized. ... Forms of electromagnetic radiation.

(from google)

thank you :)

You might be interested in

The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont

ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

$T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}$

But heat capacity of object B is twice that of object A

$T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K$

Therefore, the final temperature of both objects is 400 K.

4 0

2 years ago

A 1.2 kg ball moving due east at 40 m/s strikes a stationary 6.0 kg object. The 1.2 kg ball rebounds to the west at 25 m/s. What

RSB [31]

V_2' = v_1 + v_1'
So v_2' = 40 + -25
We have set east to be + and west -
Which gives us 15 m/s. So thats how fast the 6 kg object is going.
This is true for an elastic collision.

4 0

2 years ago

In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

Vesna [10]

Answer:

Explanation:

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

r=distance between point of rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

$v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

$v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}$

6 0

2 years ago

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

raketka [301]

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

8 0

2 years ago

A particular metal has a work function of 1.05 eV. A light is shined onto this metal with a corresponding wavelength of 324 nm.

LenaWriter [7]

Answer:

Velocity of electron will be $v=0.986\times 10^6m/sec$