• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

Messages
20
Reaction score
20
Points
13
Hello zain,

o/n 11 p43 q7 bii 2

Cannot find this question

m/j 12 p42 q1 cii
c) ii) First off, were told that the radius of the orbit of the moon increases by 4 cm each year.
We need the energy change in one year
We also know that Energy change= m.Change in phi ,where m is the mass and change in phi is change in potential.
Moon mass: 7.4 x 10^22 Kg
Change in potential= GM(1/R1 - 1/R2)
Now just substitue: Change in potential= (6.67 x 10^-11)(6 x 10^24) (1/3.84x10^8) - 1/3.84x10^8 + 0.04)= 1.09 x 10^-4
Now Energy change = moon mass x change in potential = (7.4 x 10^22)(1.09 x 10^-4)=8 x 10^18 J

o/n 12 p43 q5 c ii

Cannot find this question

o/n 11 p43 q7 bii 2
http://studentbounty.com/pastpapers...el/Physics (9702)/2011 Nov/9702_w11_qp_43.pdf

o/n 12 p43 q5 c ii
http://studentbounty.com/pastpapers...el/Physics (9702)/2012 Nov/9702_w12_qp_43.pdf
 
Messages
216
Reaction score
148
Points
53

O/n 11 p43 q7 bii 2

We need a value for plancks constant, we have to find the gradient of the graph fig. 7.1.
I will take these two points on the graph (1.3x10^-19,2.6x10^6) and (3.7x10^-19,3.8x10^6)
Gradient = y2-y1/x2-x1= (3.8-2.6)x10^6/(3.7-1.3)x10^-19= 5x10^24
But we know that the MaxEnergy of a photoelectron= hf=hc/lambda - phi

Notice that normally we plot max energy on y axis and frequency or 1/lambda on x axis. With gradient=hc

But here, they're swapped max energy of photoelectron on x axis and 1/lambda on y axis and so the gradient must equal 1/hc.

We can now continue calculating gradient =1/hc and h=1/gradient.c
So h= 1/(5x10^24)(3x10^8)= 6.67 x10^-34 Js

O/n 12 p43 q5 cii
Hmm I'm honestly not sure about how to sketch this one I don't want to give you a wrong answer
 
Messages
20
Reaction score
1
Points
3
Can anyone explain how to solve the last part of paper 3 question 1? For example part f of variant 34 M/J 2010.
 
Messages
20
Reaction score
20
Points
13
O/n 11 p43 q7 bii 2

We need a value for plancks constant, we have to find the gradient of the graph fig. 7.1.
I will take these two points on the graph (1.3x10^-19,2.6x10^6) and (3.7x10^-19,3.8x10^6)
Gradient = y2-y1/x2-x1= (3.8-2.6)x10^6/(3.7-1.3)x10^-19= 5x10^24
But we know that the MaxEnergy of a photoelectron= hf=hc/lambda - phi

Notice that normally we plot max energy on y axis and frequency or 1/lambda on x axis. With gradient=hc

But here, they're swapped max energy of photoelectron on x axis and 1/lambda on y axis and so the gradient must equal 1/hc.

We can now continue calculating gradient =1/hc and h=1/gradient.c
So h= 1/(5x10^24)(3x10^8)= 6.67 x10^-34 Js

O/n 12 p43 q5 cii
Hmm I'm honestly not sure about how to sketch this one I don't want to give you a wrong answer
http://studentbounty.com/pastpapers...el/Physics (9702)/2011 Nov/9702_w11_qp_43.pdf
can u explain q7 b ii 1 also?
 
Messages
216
Reaction score
148
Points
53

In this question were asked to calculate phi the work function energy using the eqaution we wrote in (i) and Fig 7.1
The maximum kinetic energy of an emitted photoelectron is Emax= hc/lambda - phi
Now go back to Fig 7.1 and pick any point on the line
I'll pick (1.5 x 10^-19,2.7x10^6)
h the Planck constant= 6.63 x 10^-34 Js
Just substitue and work for phi
1.5x10^-19=(6.63x10^-34)(3x10^8)(2.7x10^6) - phi
So work function of metal =Phi= 3.87 x 10^-19 J
 
Messages
20
Reaction score
20
Points
13
In this question were asked to calculate phi the work function energy using the eqaution we wrote in (i) and Fig 7.1
The maximum kinetic energy of an emitted photoelectron is Emax= hc/lambda - phi
Now go back to Fig 7.1 and pick any point on the line
I'll pick (1.5 x 10^-19,2.7x10^6)
h the Planck constant= 6.63 x 10^-34 Js
Just substitue and work for phi
1.5x10^-19=(6.63x10^-34)(3x10^8)(2.7x10^6) - phi
So work function of metal =Phi= 3.87 x 10^-19 J
but as we have to calculate 'h' in next part i don't think that we can use it in this part!
 
Messages
216
Reaction score
148
Points
53
but as we have to calculate 'h' in next part i don't think that we can use it in this part!
No bro I understand what you're trying to say but we can use h Planck's constant because we know it, and it's given in the formula sheet h= 6.63 x 10^-34 Js
In part (ii) we're asked to find a value for it using Fig 7.1 and our equation. we cool?
 
Messages
20
Reaction score
20
Points
13
No bro I understand what you're trying to say but we can use h Planck's constant because we know it, and it's given in the formula sheet h= 6.63 x 10^-34 Js
In part (ii) we're asked to find a value for it using Fig 7.1 and our equation. we cool?
k
 
Messages
216
Reaction score
148
Points
53
Can someone please explain this question?
Hey, the fundamental charge ( elementary charge) is the multiple of this set of charges of oil. And thus, the elementary charge (1.6 x10^-19 C) is a multiple of the differences of these oil charges. Subtract the first two for example,
6.4x10^ -19 - 3.2x10^-19 = 3.2 x10^-19 C
This difference is a multiple of 1.6 x 10^-19 C (the elementary charge)
 
Top