The Question is from neet-physics dual-nature-matter-radiation , and it was created by the author SUBHANI

Question

A metal surface with work function $\phi$ is illuminated by monochromatic light of frequency $
u$.  If the stopping potential is $V_0$, and the mass of an electron is $m$, the maximum speed of the emitted photoelectrons is:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Photoelectric Effect
	**Related Topics**: Work Function, Stopping Potential, Kinetic Energy of Photoelectrons
	**Prerequisites**: Understanding of energy conservation, the photoelectric effect equation, and the relationship between stopping potential and kinetic energy.
## SOLUTION PATHWAY
**Given Information**:
	* Work function of the metal surface: $\phi$
	* Frequency of the incident light: $
u$
	* Stopping potential: $V_0$
	* Mass of an electron: $m$
	* The maximum kinetic energy of the photoelectrons is equal to the work done by the stopping potential.
**Method & Steps**:
	1.	Relate the energy of the incident photon to the work function and the maximum kinetic energy of the emitted electrons.
		Reasoning: According to the photoelectric effect, the energy of the incident photon is used to overcome the work function and provide kinetic energy to the emitted photoelectrons. The equation is $E = h
u = \phi + KE_{max}$.
		Formula/Rule: $h
u = \phi + KE_{max}$
	2.	Relate the maximum kinetic energy of the photoelectrons to the stopping potential.
		Working: The maximum kinetic energy of the photoelectrons is equal to the work done by the stopping potential, which is $KE_{max} = eV_0$ where 'e' is the charge of an electron. However, the stopping potential is not directly used in this question and we will use the kinetic energy formula instead.
		Key Point: We are looking for the maximum speed, not related to stopping potential directly in this question
	3.	Express the maximum kinetic energy in terms of the maximum speed and then solve for the speed.
		Result: $KE_{max} = \frac{1}{2}mv^2_{max}$ which gives  $\frac{1}{2}mv^2_{max} = h
u - \phi$.  Solving for $v_{max}$, we get $v_{max} = \sqrt{\frac{2(h
u - \phi)}{m}}$
		Verification: The result matches option B.
 ## OPTION ANALYSIS
[A] :x: This option incorrectly adds the work function to the photon energy instead of subtracting it. It uses the formula as $KE = h
u + \phi$, instead of $KE=h
u - \phi$
[B] ✓: This option correctly calculates the maximum kinetic energy as the difference between the photon energy and the work function, and then correctly calculates the maximum speed. The formula used is $v_{max} = \sqrt{\frac{2(h
u - \phi)}{m}}$
[C] :x: This option incorrectly omits the factor of 2 in the numerator when relating kinetic energy to speed. The formula is incorrect as it should be $v_{max} = \sqrt{\frac{2(h
u - \phi)}{m}}$ not $v_{max} = \sqrt{\frac{(h
u - \phi)}{m}}$
[D] :x: This option does not have the correct dimensions for speed and also omits the square root. The formula is incorrect as it should be $v_{max} = \sqrt{\frac{2(h
u - \phi)}{m}}$ not $v_{max} = \frac{2(h
u - \phi)}{m}$
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly adding the work function to the photon energy instead of subtracting it. This is a mistake in applying the photoelectric equation. Always remember that the work function is the energy required to free the electron from the metal, so it should be subtracted from the photon's energy. 
		2.[Error 2]: Forgetting the factor of 2 or the square root when relating kinetic energy to speed. Remember that kinetic energy is given by $\frac{1}{2}mv^2$, and this has to be correctly used when solving for the speed.
	**Quick Tips**:
		1.[Helpful shortcut]:  Remember the photoelectric equation as: $h
u = \phi + KE_{max}$ and then derive the speed from the kinetic energy equation.
		2.[Memory aid]:  Use the mnemonic 'Photon energy pays for work and kinetic energy' to remember that photon energy is the sum of the work function and kinetic energy.

SUBHANI · Answer

$\sqrt{\frac{2(h
u + \phi)}{m}}$

SUBHANI · Answer

The correct Option is "$\sqrt{\frac{2(h
u - \phi)}{m}}$"

SUBHANI · Answer

$\sqrt{\frac{h
u - \phi}{m}}$

SUBHANI · Answer

$\frac{2(h
u - \phi)}{m}$

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