V is plotted against 1 lambda where lambda is wavelength of incident radiations for two metals

Answer

Text Solution

Answer : (i) `5000Å, 2000Å`, <br> (ii) `2:5` (ii) Metal-1

Solution : (i) For metal 1 threshold wavelength is `lambda_(1)` <br> `1/lambda_(1)=0.002 nm^(-1)` or `lambda_(1)=500 nm=5000 Å` <br> For metal 2 threshold wavelength is `lambda_(2)` <br> `1/lambda_(2)=0.005 nm^(-1)` or `lambda_(2)=200 nm=2000 Å` <br> (ii) `phi=(hc)/lambda_(0)` or `phi prop 1/lambda_(0)" ":. phi_(1) : phi_(2)=2 : 5` <br> (iii) Metal 1 because `lambda_(1)` lies in visible wavelength range.

Text Solution

`phi_1:phi_2:phi_3=1:2:4``phi_1:phi_2:phi_3=4:2:1``tanthetaprop((hc))/(e)`Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only

Answer : A::C

Solution :  `phi_a:phi_2:phi_3=eV_(01):eV_(02):eV_(03)` <br> `V_(01):V_(02):V_(03)=0.001:0.002:0.003=1:2:4` <br> therefore, option (a) is correct. <br> By Einstein's photoelectric equation, `(hc)/(lamda)-phi=eV` <br> `impliesV=(hc)/(elamda)-(phi)/(e)` ..(i) <br> Comparing Eq. (i) by `y=mx+c`, we get the slope of the line `m=(hc)/(e)=tantheta` <br> `implies` Option (c ) is correct. <br> From the graph it is clear that, <br> `(1)/(lamda_(01))=0.001nm^(-1)` <br> `implieslamda_(01)=(1)/(0.001)=1000nm` <br> Also, `(1)/(lamda_(02))=0.002nm^(-1)implieslamda_(02)=500nm` <br> and `lamda_(03)=250nm` <br> Violet color light will have wavelength less that 400 nm. Therefore, this light will be unable to show photoelectric effect on plate 3 `implies` Option (d) is wrong.

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1. Metal 1 may be gold and metal 2 may be cesium
2. θ1 > θ2, if metal -1 is gold and metal -2 is cesium 
3. θ1 = θ2 for any two metals
4. θ1 > θ2, if metal -1 and metal -2 are gold and copper respectively