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        . An electron accelerated through a potertial of 10,000 V ftom rest has a de-Broglie wave length 'I'. What should be the accelerEting potential so that the wave length is doubled ? (A) 20,000 v (B) 40,000 v (c) 5,000 v (D) 2500 v
2 months ago

Arun
23785 Points
							potential difference , V = 10000 volts charge on electron , q = 1.6 × 10^-19 C mass of electron, m = 9.1 × 10^-31 Kg a/c to De-broglie's wavelength, \lambda=\frac{h}{\sqrt{2qVm}} where h is plank's constant. i.e., h = 6.64 × 10^-34 Js so, wavelength = 6.64 × 10^-34/√(2 × 1.6 × 10^-19 × 10000 × 9.1 × 10^-31) = 6.64 × 10^-34/(5.4 × 10^-23) m = (6.64/5.4) × 10^(-34 + 23) m = 1.2296 × 10^-11 m = 12.296 × 10^-12 m or 12.296 pm hence, De-broglie's wavelength is 12.296 pm

2 months ago
Vikas TU
10495 Points
							Dear student The above ans is incorrect , Please do not follow the solution , Feel free to ask questions......Good Luck

2 months ago
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### Course Features

• 70 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions