SATHEE: Modern Physics 1 Question 2

Modern Physics 1 Question 2

2. An excited $H e^{+}$ ion emits two photons in succession, with wavelengths $108.5 n m$ and $30.4 n m$ , in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $λ$ , energy $E = \frac{1240 e V}{λ (in n m)}]$

(a) $n = 4$

(b) $n = 5$

(d) $n = 6$

(Main 2019, 12 April I)

radiation of wavelength $λ$ . When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $λ$ ?

(Main 2019, 10 April II) [Take, $h = 6.63 \times 10^{- 34} J s; c = 3 \times 10^{8} m s^{- 1}$ ]

(a) $9.4 n m$

(b) $12.3 n m$

(d) $11.4 n m$

Show Answer

Solution:

Change in energy in transition from $n$ to $m$ stage is given by $(n > m)$ ,

$E_{n} = - \frac{E_{0} Z^{2}}{n^{2}}$

Here, $Z = 2$

$Δ E_{n} = + 13.6 \times 4 \frac{1}{m^{2}} - \frac{1}{n^{2}} = \frac{h c}{λ}$

Let it start from $n$ to $m$ and then $m$ to ground.

So, in first case,

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Modern Physics 1 Question 2

2. An excited $H e^{+}$ ion emits two photons in succession, with wavelengths $108.5 n m$ and $30.4 n m$ , in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $λ$ , energy $E = \frac{1240 e V}{λ (in n m)}]$

Solution:

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चिकित्सा तैयारी

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Modern Physics 1 Question 2

2. An excited He+ion emits two photons in succession, with wavelengths 108.5nm and 30.4nm, in making a transition to ground state. The quantum number n corresponding to its initial excited state is [for photon of wavelength λ, energy E=1240eVλ( in nm)]

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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2. An excited $H e^{+}$ ion emits two photons in succession, with wavelengths $108.5 n m$ and $30.4 n m$ , in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $λ$ , energy $E = \frac{1240 e V}{λ (in n m)}]$