Physics 30 Matteri would apresiate some help on this question

  • Thread starter Thread starter cathoderay
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around solving a physics problem related to x-rays striking a free electron and calculating the maximum velocity of an emitted photoelectron. The user has attempted to use various formulas, including energy calculations and frequency conversions, but is confused about the final steps. They specifically question the use of the formula for kinetic energy and seek clarification on their calculations. The conversation highlights the importance of understanding energy conservation and the correct application of physics principles in solving the problem. Assistance is requested to ensure the calculations lead to the correct answer for the maximum velocity of the emitted photoelectron.
cathoderay
Messages
16
Reaction score
0
[SOLVED] physics 30 ...Matter..i would apresiate some help on this question ..

hi this is a cuestion from physics 30 grde 12 ...i need some help figuring out how to solve it ...the provlem is as folow..
1. x-rays of wavelenght 5.0 x 10^-11m strike a free electron at rest. The free electron rebounds with a velocity of 2.0 x 10^6 m/s. The Scattered X-ray strikes a metallic surface whose work function is 5.2 x 10^-16J.


What is the maximum velocity of an emitted photoelectron?
a)2.0 x 10^6 m/s
b)8.7 x 10^7 m/s
c)4.3 x 10^4 m/s
d)5.4 x 10^5 m/s


3. this is what i have done but I am just confuse now ..im not sure if this is right:

Ek=hf-hf'
f'=w/h
=5.2 X 10^-16/6.63 x 10^-34
=7.85 x 10^17HZ

E=hc/λ
=(6.63 x 10^-34)(3.00 x 10^8)/5.0 x 10^-11m
=3.978 x 10^-15J

f=E/h
=3.978 x 10^-15/6.63 x 10^-34
=6 x 10^18HZ

Ek=hf-hf'
=(6.63 x 10^-34)(6 x 10^18) - (6.63 x 10^-34)( 7.85 x 10^17)
=3.46 x 10^-15J

v=Ek/q
=3.46 x 10^-15J/1.60 x 10^-19c
=2.2 X 10^4

there I am stuck I am not sure what else to do ..what formula was i supose to use..am i in the right way or not..? if there is someone that could help me solve this problem... it will be grate ..thanks..
 
Physics news on Phys.org
cathoderay said:

v=Ek/q
=3.46 x 10^-15J/1.60 x 10^-19c
=2.2 X 10^4


what is this last step?

use E_k = \frac{1}{2} m v^2 for electron is ok here as gamma (from relativity) is only about 1.
 
Thanks..
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top