Question

Match List-I with List-II.

Choose the correct answer from the options given below:

• A. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
• B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
• C. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
• D. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Hint:

Remember these standard graphs: \[ I_p \propto \text{Intensity} \] \[ V_0=\frac{h}{e}\nu-\frac{\phi}{e} \] \[ \lambda=\frac{h}{p} \] Photoelectric current vs anode potential always shows a saturation region.

Answer 1

The Correct Option is B

Concept: Different physical quantities follow characteristic graphical relationships.

Step 1:Match photoelectric current with intensity. Photoelectric current is directly proportional to the intensity of incident radiation. \[ I_p \propto \text{Intensity} \] Hence the graph is a straight line passing through the origin. \[ (A)\rightarrow(IV) \]

Step 2: Match stopping potential with frequency. Einstein's photoelectric equation is \[ eV_0=h\nu-\phi \] or \[ V_0=\frac{h}{e}\nu-\frac{\phi}{e} \] This represents a straight line with a threshold frequency intercept. \[ (B)\rightarrow(I) \]

Step 3: Match photoelectric current with anode potential. As anode potential increases, more photoelectrons are collected and the current gradually reaches saturation. Therefore the graph rises and then becomes constant. \[ (C)\rightarrow(II) \]

Step 4: Match de-Broglie wavelength with momentum. According to de-Broglie relation, \[ \lambda=\frac{h}{p} \] Thus wavelength varies inversely with momentum. The graph is a decreasing rectangular hyperbola. \[ (D)\rightarrow(III) \]

Step 5: Write the final matching. \[ (A)\rightarrow(IV) \] \[ (B)\rightarrow(I) \] \[ (C)\rightarrow(II) \] \[ (D)\rightarrow(III) \] \[ { (A)-(IV),\ (B)-(I),\ (C)-(II),\ (D)-(III) } \] Hence, the correct option is \[ {(B)} \]