Question

From the graph, the work functions of A and B are \((h\) is Planck's constant and \(e\) is the electronic charge\().\)

• A. \(\nu_1\) and \(\nu_2\)
• B. \(V_1\) and \(V_2\)
• C. \(h\nu_1\) and \(h\nu_2\)
• D. \(\dfrac{h\nu_1}{e}\) and \(\dfrac{h\nu_2}{e}\)
• A. greater for metal A because it has a smaller work function.
• B. greater for metal B because it has a larger work function.
• C. greater for metal B because it has higher threshold frequency.
• D. the same for both metal A and metal B because it is independent of work functions of metals.
• A. the slope of the parallel lines will increase.
• B. the slope of the parallel lines will decrease.
• C. the threshold frequencies for both A and B will decrease.
• D. the slope of the parallel lines will not change but more electrons will be emitted per second.
• A. \(\dfrac13\)
• B. \(\dfrac12\)
• C. \(\dfrac25\)
• D. \(\dfrac34\)
• A. \(me\)
• B. \(\dfrac{1}{me}\)
• C. \(\dfrac{m}{e}\)
• D. \(\dfrac{e}{m}\)

Answer 1

The Correct Option is C

Concept: Einstein's photoelectric equation is \[ h\nu=\phi+K_{\max} \] where \[ \phi=h\nu_0 \] is the work function of the metal and \(\nu_0\) is its threshold frequency. The stopping potential is related to maximum kinetic energy by \[ eV_s=K_{\max}. \] Combining these equations, \[ V_s=\frac{h}{e}\nu-\frac{\phi}{e}. \] Thus, the graph of \(V_s\) versus \(\nu\) is a straight line.

Step 1: Identify threshold frequencies. The threshold frequency is obtained where \[ V_s=0. \] The corresponding intercepts on the frequency axis are \[ \nu_1 \quad \text{and} \quad \nu_2. \]

Step 2: Calculate work functions. Since \[ \phi=h\nu_0, \] the work functions of metals A and B are \[ \phi_A=h\nu_1 \] and \[ \phi_B=h\nu_2. \] Therefore, \[ \boxed{\text{Correct Option (C)}} \]

Answer 2

According to Einstein's equation, \[ K_{\max}=h\nu-\phi. \] For the same incident frequency \(\nu\), \[ K_{\max} \] depends upon the work function.

Step 1: Compare work functions. From the graph, \[ \nu_2>\nu_1. \] Therefore, \[ \phi_B=h\nu_2 > h\nu_1=\phi_A. \] Hence metal B has a larger work function.

Step 2: Compare maximum kinetic energies. Since \[ K_{\max}=h\nu-\phi, \] a smaller work function produces a larger kinetic energy. Thus, \[ K_{\max}(A) > K_{\max}(B). \] Therefore, \[ \boxed{\text{Correct Option (A)}} \]

Answer 3

Concept: The photoelectric equation is \[ V_s=\frac{h}{e}\nu-\frac{\phi}{e}. \] The slope is \[ \frac{h}{e}. \] Since \(h\) and \(e\) are constants, the slope does not depend upon intensity.

Step 1: Effect of increasing intensity. Increasing intensity increases the number of incident photons per second. Hence more electrons are emitted per second.

Step 2: Effect on stopping potential and threshold frequency. Stopping potential depends on frequency and not on intensity. Threshold frequency is a characteristic property of the metal and remains unchanged. Therefore, \[ \boxed{\text{Correct Option (D)}} \]

Answer 4

Using Einstein's equation, \[ K_{\max}=h(\nu-\nu_0). \]

Step 1: Calculate \(E_1\). For frequency \[ \nu=3\nu_0, \] \[ E_1=h(3\nu_0-\nu_0). \] \[ E_1=2h\nu_0. \]

Step 2: Calculate \(E_2\). For frequency \[ \nu=6\nu_0, \] \[ E_2=h(6\nu_0-\nu_0). \] \[ E_2=5h\nu_0. \]

Step 3: Find ratio. \[ \frac{E_1}{E_2} = \frac{2h\nu_0}{5h\nu_0}. \] \[ \boxed{ \frac{E_1}{E_2} = \frac25 } \] Hence, \[ \boxed{\text{Correct Option (C)}} \]

Answer 5

The stopping potential equation is \[ V_s=\frac{h}{e}\nu-\frac{\phi}{e}. \] Comparing with the equation of a straight line, \[ y=mx+c, \] the slope is \[ m=\frac{h}{e}. \] Multiplying both sides by \(e\), \[ h=me. \] Therefore, \[ \boxed{h=me} \] Hence, \[ \boxed{\text{Correct Option (A)}} \]