Quantum Mechanics

Step 1: First, we need to calculate the energy of the photon using the equation:
$E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D$

Where $h$ is Planck’s constant, $c$ is the speed of light, and $%5Clambda$ is the wavelength of the incident light.
Step 2: For the given wavelength of $%5Clambda%20%3D%204770%20%5C%2C%20%5Ctext%7B%C3%85%7D$ , we can calculate the energy of the photon as:
$E%20%3D%20%5Cfrac%7B(6.626%20%5Ctimes%2010%5E%7B-34%7D%20%5C%2C%20%5Ctext%7BJs%7D)%20(3%20%5Ctimes%2010%5E8%20%5C%2C%20%5Ctext%7Bm%2Fs%7D)%7D%7B4770%20%5Ctimes%2010%5E%7B-10%7D%20%5C%2C%20%5Ctext%7Bm%7D%7D%20%3D%204.15%20%5C%2C%20%5Ctext%7BeV%7D$
Step 3: Electrons will be emitted from a metal surface only if the energy of the incident photon is greater than or equal to the work function ( $%5Cphi$ ) of the metal.
Step 4: Comparing the photon energy with the work functions, we find that only metal D (with a work function of $2.3%20%5C%2C%20%5Ctext%7BeV%7D$ ) will emit electrons because its work function is less than or equal to the energy of the photon.

Conclusion: Metal D will emit electrons as its work function ( $2.3%20%5C%2C%20%5Ctext%7BeV%7D$ ) is less than the photon energy ( $4.15%20%5C%2C%20%5Ctext%7BeV%7D$ ).

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Quantum Mechanics

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs