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Julli [10]
2 years ago
12

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

kinetic energy of the electrons when radiation of wavelength 198nm is used? c. It is observed that the electrons may be prohibited from reaching the anode by applying a stopping potential. What is the stopping potential in the case when radiation of wavelength 198nm is used?
Physics
1 answer:
Tanya [424]2 years ago
5 0

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

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A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi
Jobisdone [24]
At the top:

         Potential Energy = (mass) x (gravity) x (height)

                                       = (30 kg) x (9.8 m/s²) x (3 meters)

                                       =      882 joules

At the bottom:

           Kinetic Energy  =  (1/2) x (mass) x (speed)²

                                       = (1/2) x (30 kg) x (3 m/s)²

                                       =        (15 kg)  x  (9 m²/s²)

                                       =              135 joules .

He had  882 joules of potential energy at the top,
but only  135 joules of kinetic energy at the bottom.

Friction stole  (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
6 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
A seagull flying horizontally at 8.00m/s carries a clam with a mass of 300g in its beak. Calculate the total mechanical energy o
Stells [14]

Answer:

9.6J+88.2J=97.8J

Explanation:

Here the velocity of the seagull is given,mass is given and its height.

We have to find its mechanical energy my friend.

Mechanical energy=kinetic energy + potential energy.

First we will find kinetic energy.

For calculating kinetic energy we need mass and velocity,which are given here.

So, Ek=

1 \div 2mv {?}^{2}

So by substituting the values we get 9.6J.

Now we find the potential energy which is mgh.

By substituting the values we get 88.2J.

Then we add both of those and get 97.8J

I hope this satisfies you and make sure you contact me if it doesn't

7 0
3 years ago
Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

                           w=\frac{2*wavelength*d}{a}

where, d is the distance between slit and the screen, and a is the slit width.

  • Thus, distance between slit and the screen is,

                           d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0
1 year ago
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Please look at the attached image below for the explanation

5 0
2 years ago
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