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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Dual Nature of Matter

JIPMER 2016 · Physics · Dual Nature of Matter

The graph (1)/(λ) and stopping potential (V) of three metals having...

The graph 1λ\frac{1}{\lambda} and stopping potential (V)(V) of three metals having work function ϕ1,ϕ2\phi_1, \phi_2 and ϕ3\phi_3 in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is/are correct? [Here λ\lambda is the wavelength of the incident ray] The graph (1)/(λ) and stopping potential (V) of three metals having work function φ_1, φ_2 and φ_3 in an experiment of (i) Ratio of work functions ϕ1:ϕ2:ϕ3=1:2:4\phi_1: \phi_2: \phi_3=1: 2: 4 (ii) Ratio of work functions ϕ1:ϕ2:ϕ3=4:2:1\phi_1: \phi_2: \phi_3=4: 2: 1 (iii) tan⁡θ∝hce\tan \theta \propto \frac{h c}{e}, where h=h= Planck's constant, c=c= speed of light (iv) The violet colour-light can eject photoelectrons from metals 2 and 3

Options

  1. A(i), (iii)
  2. B(i), (iv)
  3. C(ii), (iii)
  4. D(i), (ii) and (iv)

Correct answer

A. (i), (iii)

Step-by-step solution

According to Einstein photoelectric equation, E=Kmax⁡+ϕ∴hcλ=ϕ+eVS∴VS=[hcλ−ϕ]1e\begin{aligned} E & =K_{\max }+\phi \\ \therefore \quad \frac{h c}{\lambda} & =\phi+e V_S \\ \therefore \quad V_S & =\left[\frac{h c}{\lambda}-\phi\right] \frac{1}{e}\end{aligned}V_Sversusversus\frac{1}{\lambda}graphisastraightline.Slope,graph is a straight line. Slope,\tan \theta=\frac{h c}{e}ThecurvesintersecttheThe curves intersect the\frac{1}{\lambda}axis,whereaxis, whereV_Siszero.is zero.∴hcλ−ϕ=0⇒1λ=ϕhc\begin{aligned} \therefore & \frac{h c}{\lambda}-\phi & =0 \\ \Rightarrow & \frac{1}{\lambda} & =\frac{\phi}{h c}\end{aligned}FormetalFor metal1 \quadFormetal2Formetal3For metal 2 For metal 3ϕ1hc=0.001ϕ2hc=0.002ϕ3hc=0.004∴ϕ1:ϕ2:ϕ3=1:2:4\begin{array}{cc} & \frac{\phi_1}{h c}=0.001 \quad \frac{\phi_2}{h c}=0.002 \quad \frac{\phi_3}{h c}=0.004 \\ \therefore \quad & \phi_1: \phi_2: \phi_3=1: 2: 4\end{array}Formetal2,thresholdwavelength,For metal 2, threshold wavelength,\lambda O2=hcϕ=10.002=500 nmSimilarly,\mathrm{O}_2=\frac{h c}{\phi}=\frac{1}{0.002}=500 \mathrm{~nm} Similarly,\quad \lambda o3=250 nmλviolet =400 nmSo\mathrm{o}_3=250\mathrm{~nm} \lambda_{\text{violet }}=400 \mathrm{~nm} So\lambda_violet <λthreshold {\text{violet }}\lt \lambda_{\text{threshold }}formetal1,for metal 1 ,\lambda_{\text{violet }}^{\text{violet }}\gt\lambda_{$\text{threshold }} for metal 3 Thus, violet light will eject photoelectron for metal 2 and not metal 3 .
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