SATHEE: Modern Physics 1 Question 48

Modern Physics 1 Question 48

44. Consider a hydrogen atom with its electro in the $n^{th}$ orbital. An electromagnetic radiation of wavelength $90 n m$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 e V$ , then the value of $n$ is $(h c = 1242 e V n m)$ (2015 Adv.)

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Solution:

Kinetic energy of ejected electron

$=$ Energy of incident photon - energy required to ionize the electron from $n$ th orbit (all in $e V$ )

$\begin{aligned} ∴ 10.4 & = \frac{1242}{90} - | E_{n} | \\ = \frac{1242}{90} - \frac{13.6}{n^{2}} (as E_{n} \propto \frac{1}{n^{2}} and E_{1} = - 13.6 e V) \end{aligned}$

Solving this equation, we get

$n = 2$

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Modern Physics 1 Question 48

44. Consider a hydrogen atom with its electro in the nth orbital. An electromagnetic radiation of wavelength 90nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4eV, then the value of n is (hc=1242eVnm) (2015 Adv.)

Analytical & Descriptive Questions

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44. Consider a hydrogen atom with its electro in the $n^{th}$ orbital. An electromagnetic radiation of wavelength $90 n m$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 e V$ , then the value of $n$ is $(h c = 1242 e V n m)$ (2015 Adv.)