Question

An electron in a hydrogen like species is in the excited state n2. The wavelength for the transition n2 to n1 = 2 is 48.24 nm. The corresponding wavelength for the transition n2 to n1 = 3 is 142.46 nm. Find the value of n2 and z? 10. What transition in the H-spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ ion spectrum?

Answer 1

Explanation:

Answer:

n2 = 5 and Z = 3

Explanation:

Here, It is given that an electron in a hydrogen like species is in the excited state (n2). the wavelengths corresponding to a transition second orbit is 48.24nm

∴ n1 = 2 then λ = 48.24 nm

=>  from same orbit, wavelength corresponding to a transition to third orbit is 142.46 nm.

∴ n2 = 3 then  λ = 142.46 nm

=> To calculate n2 and Z we can use the below formula:

1/λ = R_H*Z² [1/n₁² - 1/n₂²]

1.48.24 = R_H*Z² R_H*Z² [1/n₁² - 1/n₂²]...(1)

1/142.46=R_H*Z² [1/9 - 1/n₂²] ...(2)

=> dividing eq (1) by (2), we get

142.46/48.24 = n₂² - 4/ n₂² - 9 * 9/4

1.31 = n₂² - 4/ n₂² - 9

n₂ = 5

=> By putting this value in equation (1) we can calculate the atomic number:

Rh = 109677 cm⁻¹

λ = 48.24 * 10⁻⁷ cm

1/48.24 * 10⁻⁷ = 109677 * Z² [1/4 - 1/25]

Z = 3

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