A particle having electric charge 3 times 10^-19 text C and mass 6 times 10^-27 text kg is accelerated by applying an electric potential of 1.21 text V. Wavelength of the matter wave associated with the particle is alpha times 10^-12 text m. The value of alpha is ________. (Take Planck's constant = 6.6 times 10^-34 text Jcdottexts)

Numerical Answer Type:
Enter a numerical value Answer: 10 to 10 +4 marks

Solution & Explanation

### Related Formula lambda = frachp = frachsqrt2mK K = qV ### Core Logic For a charged particle accelerated through a potential difference V, the de Broglie wavelength is: lambda = frachsqrt2mqV ### Step 1: Value Substitution lambda = frac6.6 times 10^-34sqrt2 times 6 times 10^-27 times 3 times 10^-19 times 1.21 lambda = frac6.6 times 10^-34sqrt36 times 10^-46 times 1.21 ### Step 2: Arithmetic Evaluation Evaluate the square root: sqrt36 times sqrt1.21 times sqrt10^-46 = 6 times 1.1 times 10^-23 = 6.6 times 10^-23 lambda = frac6.6 times 10^-346.6 times 10^-23 lambda = 10^-11 text m ### Step 3: Final Conclusion We need to express this in the form alpha times 10^-12 text m: 10^-11 text m = 10 times 10^-12 text m Therefore, alpha = 10. ### Pattern Recognition The expression inside the radical always evaluates cleanly in JEE. 2 times 6 times 3 = 36 and 1.21 = (1.1)^2 are engineered to perfectly cancel the 6.6 in the numerator. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

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