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VLD [36.1K]
3 years ago
12

As the rate of radioactive decay becomes smaller, half-life becomes ____.

Physics
1 answer:
Marina86 [1]3 years ago
8 0

Answer:

Larger

Explanation:

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A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen
leva [86]

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

  • Work function is a material property defined as the minimum amount of energy  required to infinitely remove electrons from the surface of a particular solid.
  • The potential difference required to support all emitted electrons is called the stopping potential which is given by v_0=\frac{K.E_m_a_x}{e} .....(1)
  • where v_0 is the stopping potential and e is the charge of the electron given by 1.6\times10^-^1^9 .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation  is given by:

K.E_m_a_x=E-\phi      ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

  ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV

As we know that E=\frac{hc}{\lambda}  ....(3)

where Speed of light,c = 3\times10^8 m/s and Planck's constant , h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs

From equation (3) , we get

\lambda=\frac{hc}{E} \\\\\lambda=\frac{  4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm

Learn about more einstein photoelectric equation  here:

brainly.com/question/11683155

#SPJ4

8 0
1 year ago
The position of a certain airplane during takeoff is given by x=1/2 *bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instan
Serhud [2]

Answer:

1362000 kgm/s

Explanation:

So the total mass combination of the plane and the people inside it is

M = 35000 + 160*65 = 45400 kg

After 15 seconds at an acceleration of 2 m/s2, the plane speed would be

V = 2*15 = 30 m/s

So the magnitude of the plane 15s after brakes are released is

MV = 45400 * 30 = 1362000 kgm/s

5 0
3 years ago
Tectonic plates move due to _____ forces.
wel
<span>Tectonic plates move because they are floating on top of the liquid mantle. The mantle itself moves due to convection currents: hot rock rises, gives off some heat, then falls. This creates vast swirls of moving liquid rock under the crust of the earth, which jostles the plates of crust on top.
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6 0
3 years ago
Read 2 more answers
A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =
Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

v = \omega R\\a = \alpha R

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft

One revolution is equal to the circumference of the drum. So, total number of revolutions is

x / (2\pi R) = 6/(\pi R)

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2

b) a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2

5 0
3 years ago
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
3 years ago
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