Physics: Post your doubts here!

Holmes

Holmes

Messages
438
Reaction score
3,645
Points
253
Can any physics geek here tell me about physics P5.
I don't know a single word.
Help please.

Can anyone post a single solved P5 of physics, It would be a great help.
 
Thought blocker

Thought blocker

XPRS Moderator
Messages
8,544
Reaction score
34,867
Points
718
Holmes said:
Can any physics geek here tell me about physics P5.
I don't know a single word.
Help please.

Can anyone post a single solved P5 of physics, It would be a great help.
Click to expand...
amina1300 said:
same :'(
Click to expand...

First of all, make your concepts clear of your theory paper.
Then dive into P5.

At the start, it's gonna freak you out. Be patient.

Solve 4 or 5 or 6 papers with marking scheme, try to understand how to write answers and then discuss your doubts here or on our whats app group or wid your teachers and keep on improving your skills to tackle paper 5.

Good luck folks.
 
Thought blocker

Thought blocker

XPRS Moderator
Messages
8,544
Reaction score
34,867
Points
718
If anyone who wants to join our whats app group do leave me a whats app msg on +919426116018. I am Rohan.
There are many of the people who post their doubts there and gets instant replies. (y)
 
Holmes

Holmes

Messages
438
Reaction score
3,645
Points
253
Thought blocker said:
First of all, make your concepts clear of your theory paper.
Then dive into P5.

At the start, it's gonna freak you out. Be patient.

Solve 4 or 5 or 6 papers with marking scheme, try to understand how to write answers and then discuss your doubts here or on our whats app group or wid your teachers and keep on improving your skills to tackle paper 5.

Good luck folks.
Click to expand...
Bro! what to do about diagrams ???
 
anastasia grey113

anastasia grey113

Messages
665
Reaction score
13,608
Points
503
Aishayasin said:
View attachment 63174


Part bii in detail please
View attachment 63175
Click to expand...
okay so you have attempted part b)i) right?
that shows you know what work function is
you are half way through to the answer
the point that is to be noted is that it is the MINIMUM energy required
this will help us understand the table
on tungsten, if a wave having photons of energy lesser than 4.49 is incident, it will NOT emit any electrons
you can read off sodium and potassium's values off the table in the same way

we are given two waves of wavelengths 350nm and 700nm
v just have to figure out which metal will emit electrons when one of the waves is incident on it
for this we will have to find the energy of photons in each wave using
E = hc/ λ where h=Planck's constant, c= speed of light and λ=wavelength of wave
Now putting in the values for the first wave we have,
Wave (700nm)
E= (6.63x10-34 x 3x108)/700x10-9
the answer will be 2.84x10-19 Joules

Use the same formula for the 2nd wave
the answer will be 5.68x10-19 Joules

Now as you can see that the values we have calculated are in Joules but the work functions are in eV
so convert these values using
eV = E/1.6x10-19
This way the answers will be
first wave (700nm) = 1.78 eV
2nd wave (350nm) = 3.55 eV

Now it will be easier to analyse the values. We just have to find the metal which has the work function LESSER THAN the energy of photons in one of the waves.
The first wave is definitely not suitable for any of the three metals since all of them have work functions higher than its energy.
The 2nd wave?
Only Potassium seems to fit in because of its lower work function that the energy of 2nd wave.
SO POTASSIUM WILL EMIT ELECTRONS WHEN 350nm WAVE IS INCIDENT ON IT.

Basically, we were asked to find a pair of metal and wave in which the photon energy of the wave was greater than work function of the metal.
 
anastasia grey113

anastasia grey113

Messages
665
Reaction score
13,608
Points
503
GarryTheGhost said:
part C only please
Click to expand...
First of all, see what they are asking for: the force experienced by the wires.
And F is directly proportional to B as B=F/IL
so in order to have a high magnitude of force between the two wires, a high value of B is required.
But as you can see, the constant in the formula given is very small (2x10-7)
so even if a force is produced under normal circumstances, it will be so small that it would be easily overlooked

so for a very noticeable value of B (let's suppose 1T), I/d would have to be 1/2 x10-7 which is a very small value
and for that the current has to be very large around 5000A and the distance has to be as small as 1mm
but this is practically impossible so you see this ratio can never be achieved

and even if it is, the force produced might not exactly produce the same effect because the wires also have weight which acts in a perpendicular direction.
so overall resultant will be downwards as weight of wire must be very large (let's say 0.2N) but the relative force must be very small with a usual 220V ac current and a distance of around minimum 1cm.
 
Last edited:
You must log in or register to reply here.
Top