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NEET PHYSICSDUAL NATURE OF RADIATION AND MATTEREasy

Question

The dependence of the short wavelength limit λmin\lambda_{min} on the accelerating potential VV is represented by the curve of figure:

A

Curve A

B

Curve B

C

Curve C

D

None of these

Step-by-Step Solution

The short wavelength limit (λmin\lambda_{min}), also known as the cutoff wavelength, for X-rays produced in a tube is determined by the maximum energy of the incident electrons. The entire kinetic energy (K=eVK = eV) of an electron is converted into a single photon of maximum energy (Emax=hνmaxE_{max} = h\nu_{max}). Using the relation E=hcλE = \frac{hc}{\lambda}, we get: eV=hcλmineV = \frac{hc}{\lambda_{min}} λmin=hce⋅1V\lambda_{min} = \frac{hc}{e} \cdot \frac{1}{V}

Since hh, cc, and ee are constants (Source ), λmin\lambda_{min} is inversely proportional to the accelerating potential VV (λmin∝1/V\lambda_{min} \propto 1/V). The graph representing this inverse relationship is a rectangular hyperbola. Assuming standard graphical representations where curve C depicts a hyperbola (decreasing λ\lambda as VV increases), C is the correct representation.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from DUAL NATURE OF RADIATION AND MATTER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSDUAL NATURE OF RADIATION AND MATTERdependencewavelengthlambdaminacceleratingpotential

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