Question

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

Answer 1

answer : (a) 91nm , (b) 22.75 nm (c) 10.11 nm

using Rydberg's wavelength equation,

1/λ = RZ²[1/n1² - 1/n2² ]

for smallest wavelength, n2 = ∞ and n1 = 1

so, 1/λ = RZ² [1/1² - 1/∞²] = RZ²

(a) hydrogen, Z = 1

so, 1/λ = R(1)² = R

⇒λ = 1/R = 1/1.1 × 10^7 m ≈ 91 nm

(b) Helium( He+) , Z = 2

1/λ = RZ² = R(2)² =4R

⇒λ = 91/4 = 22.75 nm

(c) Lithium (Li++), Z = 3

1/λ = R(3)² = 9R

λ = 1/9R = 91/9 = 10.11 nm

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Answer 2

The smallest wavelength of radiation emitted by  a) Hydrogen is 91 nm  ,b) He++ is 23 nm ,c) Li++ is 10 nm.

Explanation:

For the smallest wavelength, The energy must be maximum and for maximum energy transition, n1 = 1 and n2 = infinity and now,

The formula to determine the wavelength is given as follows

$\frac{1}{\lambda } = Z^2R(\frac{1}{n_1^2} -\frac{1}{n_2^2} )$

We know the for Hydrogen Z=1,

R= Rydberg constant = $1.097 \times 10^{7} m^{-1}$

$\lambda=\frac{1}{1.097 \times 10}=\frac{1}{1.097} \times 10^{-7}$

$\Rightarrow\lambda=91.01 \times 10^{-9}=91 \mathrm{nm}$

Now substituting Z = 1 in the above equation we get wavelength = 91 nm  

And for Helium, Z = 2,

$\frac{1}{\lambda}=(2)^{2}\left(1.097 \times 10^{7}\right)\left(\frac{1}{\left(1^{2}\right)}-\frac{1}{\left(\infty^{2}\right)}\right)$

$\Rightarrow\lambda=\frac{91 n m}{4}=23 n m$

Now substituting Z =2 in the above equation, we get wavelength = 23 nm

And for Lithium, Z = 3,  

$\frac{1}{\lambda}=(3)^{2} \times\left(1.097 \times 10^{7}\right)\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)$

$\Rightarrow \lambda=\frac{91 n m}{Z^{2}}=\frac{91}{9}=10 n m$

Now substituting Z = 3 in the above equation, we can get wavelength = 10nm.

Thus the smallest wavelength for Hydrogen, Helium and Lithium is calculated.