The Hitomi satellite recently observed the Ly | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the Rydberg formula for hydrogen-like ions, the wavelength $\lambda$ of the emission line is given by:
$\frac{1}{\lambda} = R Z^{2} \left(

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Using the Rydberg formula for hydrogen-like ions, the wavelength $\lambda$ of the emission line is given by:
$\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} ight)$
Given:
Rydberg constant $R \approx 1.097 	imes 10^{7} \, m^{-1} \approx 1.1 	imes 10^{7} \, m^{-1}$
Atomic number of iron $Z = 26$
Initial state $n_{i} = 2$
Final state $n_{f} = 1$
Substituting the values:
$\frac{1}{\lambda} = (1.1 	imes 10^{7}) 	imes (26)^{2} 	imes \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} ight)$
$\frac{1}{\lambda} = 1.1 	imes 10^{7} 	imes 676 	imes ( 1 - 0.25 )$
$\frac{1}{\lambda} = 1.1 	imes 10^{7} 	imes 676 	imes 0.75$
$\frac{1}{\lambda} = 5.577 	imes 10^{9} \, m^{-1}$
Calculating $\lambda$:
$\lambda = \frac{1}{5.577 	imes 10^{9}} \approx 1.79 	imes 10^{-10} \, m$
Converting to $\mathring{A}$ $(1 \, \mathring{A} = 10^{-10} \, m)$:
$\lambda \approx 1.79 \, \mathring{A}$
Thus, the wavelength is closest to $2 \, \mathring{A}$.

The Hitomi satellite recently observed the Lyman alpha emission line ($n=2$ to $n=1$) of hydrogen-like iron ion (atomic number of iron is $26$) from the Perseus galaxy cluster. The wavelength of the line is closest to ............... $\mathring{A}$.

Similar Questions

It is found experimentally that $13.6 \; eV$ energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level $n$ to level $(n-1)$. For large $n$,show that this frequency equals the classical frequency of revolution of the electron in the orbit.

In Bohr's atomic model, the energy of the electron is $E$ in the second orbit of a hydrogen atom. The energy of the electron in the third orbit of a helium atom $(Z = 2)$ will be:

Magnetic moment due to the motion of the electron in the $n$th energy state of a hydrogen atom is proportional to ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module