JEE Advance - Chemistry (2023 - Paper 2 Online - No. 11)
For $\mathrm{He}^{+}$, a transition takes place from the orbit of radius $105.8 \mathrm{pm}$ to the orbit of radius $26.45 \mathrm{pm}$. The wavelength (in nm) of the emitted photon during the transition is _______.
[Use :
Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$
Rydberg constant, $R_{\mathrm{H}}=2.2 \times 10^{-18} \mathrm{~J}$
Planck's constant, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
Speed of light, $\mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ ]
[Use :
Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$
Rydberg constant, $R_{\mathrm{H}}=2.2 \times 10^{-18} \mathrm{~J}$
Planck's constant, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
Speed of light, $\mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ ]
Answer
30
Explanation
1. The radius of the nth orbit in a single-electron system is given by the equation:
$$r = \frac{52.9 \times n^2}{Z}$$
where $r$ is in pm, $n$ is the principal quantum number, and $Z$ is the atomic number.
2. For a $\mathrm{He}^{+}$ ion, $Z=2$.
3. We are given the initial radius $r_2 = 105.8$ pm and the final radius $r_1 = 26.45$ pm.
4. We can substitute these radii into the equation for $r$ to find the corresponding quantum numbers.
For $r_2 = 105.8$ pm, we get :
$$105.8 = \frac{52.9 \times n_2^2}{2}$$
which gives $n_2 = 2$.
Similarly, for $r_1 = 26.45$ pm, we get :
$$26.45 = \frac{52.9 \times n_1^2}{2}$$
which gives $n_1 = 1$.
5. Therefore, the transition is from $n_2 = 2$ to $n_1 = 1$.
6. The energy difference during this transition is equal to the energy of the emitted photon, which is given by :
$$E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$
where $h$ is the Planck's constant, $c$ is the speed of light, $R_H$ is the Rydberg constant, and $\lambda$ is the wavelength of the photon.
7. Substituting all known values into this equation gives :
$$\frac{6.6 \times 10^{-34} \, \mathrm{J} \, \mathrm{s} \times 3 \times 10^8 \, \mathrm{m/s}}{\lambda} = 2.2 \times 10^{-18} \, \mathrm{J} \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$$
Solving this equation yields $\lambda = 30 \times 10^{-9}$ m, or 30 nm.
8. So, the wavelength of the emitted photon during the transition is 30 nm.
$$r = \frac{52.9 \times n^2}{Z}$$
where $r$ is in pm, $n$ is the principal quantum number, and $Z$ is the atomic number.
2. For a $\mathrm{He}^{+}$ ion, $Z=2$.
3. We are given the initial radius $r_2 = 105.8$ pm and the final radius $r_1 = 26.45$ pm.
4. We can substitute these radii into the equation for $r$ to find the corresponding quantum numbers.
For $r_2 = 105.8$ pm, we get :
$$105.8 = \frac{52.9 \times n_2^2}{2}$$
which gives $n_2 = 2$.
Similarly, for $r_1 = 26.45$ pm, we get :
$$26.45 = \frac{52.9 \times n_1^2}{2}$$
which gives $n_1 = 1$.
5. Therefore, the transition is from $n_2 = 2$ to $n_1 = 1$.
6. The energy difference during this transition is equal to the energy of the emitted photon, which is given by :
$$E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$
where $h$ is the Planck's constant, $c$ is the speed of light, $R_H$ is the Rydberg constant, and $\lambda$ is the wavelength of the photon.
7. Substituting all known values into this equation gives :
$$\frac{6.6 \times 10^{-34} \, \mathrm{J} \, \mathrm{s} \times 3 \times 10^8 \, \mathrm{m/s}}{\lambda} = 2.2 \times 10^{-18} \, \mathrm{J} \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$$
Solving this equation yields $\lambda = 30 \times 10^{-9}$ m, or 30 nm.
8. So, the wavelength of the emitted photon during the transition is 30 nm.
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