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Academic session 2026-27, Chapter 2 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Structure of Atom. This page offers a step-by-step solution to the specific question from Exercise 1, Question 61:
If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.
From Heisenberg’s uncertainty principle,

Where,
Δx = uncertainty in position of the electron
Δp = uncertainty in momentum of the electron
Δx = 0.002nm = 2x10-12m(given)
Therefore, Substituting the values in the expression of Δp:
Δp = h/4π Δx or

= 2.637 × 10–23 Jsm–1
Δp = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s–1)
Actual momentum = h/4πx 0.05nm
= 6.626x10-34/4 x3.14 x 5 x10-11
= 1.055 x 10-24 kg m/sec
Following results are observed when sodium metal is irradiated with different wavelengths.
Calculate (a) threshold wavelength and, (b) Planck’s constant.
| λ (nm) | 500 | 450 |
400 |
| v × 10–5 (cm s–1) | 2.55 | 4.35 | 5.35 |
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 61: If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertai....
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Where is the given value (0.05) in formula
This question is not asked.
Is there is any problem in defining the value