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QPages 384-386: Q3: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q3 Solution: Calculate the smallest wavelength of radiation that may be…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often think the smallest wavelength comes from the first allowed transition (e.g. $n=2 o1$). In reality the minimum wavelength is obtained in the limit $n oinfty o1$, so use $1/lambda=R,Z^2(1/1^2-0)$ to find $lambda_{min}$.

Calculate the smallest wavelength of radiation that may be emitted by a) hydrogen, b) He+^+ and c) Li++^{++}.

Step-by-Step Verified Solution

**Answer:** (a) 91 nm (b) 23 nm (c) 10 nm. **Step-by-Step Explanation:** The smallest wavelength (highest energy) is emitted when the electron jumps from n = � to n = 1. (a) Hydrogen (Z=1): 1λ=R(1/12−0)=R\frac{1}{\lambda} = R(1/1^2 - 0) = R. λ=1/R=1/(1.097×107)=91.2\lambda = 1/R = 1/(1.097 \times 10^7) = 91.2 nm. (b) He_ (Z=2): 1λ=RZ2=4R\frac{1}{\lambda} = RZ^2 = 4R. λ=1/(4R)=91.2/4=22.8\lambda = 1/(4R) = 91.2/4 = 22.8 nm � 23 nm. (c) Li__ (Z=3): 1λ=RZ2=9R\frac{1}{\lambda} = RZ^2 = 9R. λ=1/(9R)=91.2/9=10.1\lambda = 1/(9R) = 91.2/9 = 10.1 nm � 10 nm [citation:3][citation:7].

All Questions in this Chapter

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