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Question #44189

Difficulty: 8/10 | 🟢 Free

The Question

Due to poor shielding by inner core electrons, certain transition metal groups exhibit anomalous periodic trends in their ionization energies and work functions ( $\Phi$ ). When monochromatic light of energy $6.50\text{ eV}$ is incident on the surfaces of Silver (Ag) and Gold (Au), the magnitude of the difference in their stopping potentials is observed to be $1.00\text{ V}$ . Given that the threshold wavelength for photoelectric emission from Silver is $310\text{ nm}$ , determine the threshold wavelength for Gold. (Use $hc = 1240\text{ eV}\cdot\text{nm}$ )

Options

A.
$248\text{ nm}$
B.
$413\text{ nm}$
C.
$827\text{ nm}$
D.
$354\text{ nm}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Photoelectric Effect Threshold|Periodic Properties of Lanthanides|Post-Lanthanoid Radii Equality

Why This Question Is Tricky

Common Mistakes

The question tests the intersection of the Photoelectric Effect and Periodic Properties.

The 'Hidden Variable' is the Lanthanoid contraction and relativistic stabilization of the 6s orbital in 5d transition metals (like Gold).

Students relying on the general textbook rule that metallic character increases (and work function decreases) down a group will incorrectly assume $\Phi_{Au} < \Phi_{Ag}$ .

This leads them to subtract $1.00\text{ eV}$ from Silver's work function instead of adding it, trapping them with a calculated threshold wavelength of $413\text{ nm}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Photoelectric Effect Threshold
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free