A

1. 13.6 Ev To Hz
2. 13.6 Ev Binding Energy
3. 13.6 Ev Hydrogen
4. 13.6 Ev To Joules
5. 13.6 Ev To Erg

Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history. 1 The ground state energy of hydrogen atom is -13 6 eV (i)What the potential and kinetic energy of an electron in the 3rd excited state (ii)if the electron jumps to the ground state from the third excited state,calculate the frequency of photon emitted - Physics - Atoms. The ground state energy of H-atom is 13.6 eV. The energy needed to ionize H-atom from its second excited state is 3:05 000+ LIKES. In relation to the base unit of energy = (joules), 1 Rydberg (Ry) is equal to 2.179872E-18 joules, while 1 Electron Volt (eV) = 1.60218E-19 joules. 1 Rydberg to common energy units. That 13.6eV is the difference in the energy of a bound electron as compared to an electron at rest far from the proton. So it is an energy difference. We don't include the rest mass because that's a constant so it's the same for both the bound and free electron. When we take the difference in the energies the rest mass cancels out.

$10.2, eV$

$3.4 ,eV$

Solution:

Energy required to remove an electron from $n$ th orbit is,
$E_{n} =- frac{13.6}{n^{2}}$
Here, $n =2$
Therefore, $E_{2}= -frac{13.6}{2^{2}}= -3.4 V$

1. The first line of the Lyman series in a hydrogen spectrum has a wavelength of $1210 Å$. The corresponding line of a hydrogen- like atom of $Z = 11$ is equal to

2. The inverse square law in electrostatics is$left|vec{F}right| = frac{e^{2}}{left(4pivarepsilon_{0}right)cdot r^{2}}$ for the force between an electron and a proton. The $(frac{1}{r})$ dependence of $|vec{F}|$ can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass $m_p$, force would be modified to
$left|vec{F}right| = frac{e^{2}}{left(4pivarepsilon_{0}right)r^{2}}left[frac{1}{r^{2}} + frac{lambda}{r}right] cdot exp left(-lambda rright)$
where $lambda = m_pc/ħ$ and $ħ= frac{h}{2pi}$ The change in the ground state energy $(eV)$ of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron. ($r_B =$ Bohr’s radius)

3. Assertion : Atoms of each element are stable and emit characteristic spectrum.
Reason : The spectrum provides useful information about the atomic structure.

4. What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series ?

5. Assertion : Atom as a whole is electrically neutral.
Reason : Atom contains equal amount of positive and negative charges.

13.6 Ev To Hz

6. If the wavelength of the first line of the Balmer series of hydrogen is $6561 , Å$, the wavelength of the second line of the series should be

7. Match the correct pairs.

Emission series		Make transitions from higher levels to following levels
A	Lyman series	P	$n = 1$
B	Paschen series	Q	$n = 2$
C	Balmer series	R	$n = 3$
D	Brackett series	S	$n = 4$
T	$n = 5$

8. Assertion :According to electromagnetic theory an accelerated particle continuously emits radiation.
Reason : According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic.

9. If $upsilon_{1}$ is the frequency of the series limit of Lyman series, $upsilon_{2}$ is the frequency of the first line of Lyman series and $upsilon_{3}$ is the frequency of the series limit of the Balmer series, then

10. Assertion :In alpha particle scattering number of alpha particle undergoing head on collision is small.
Reason : Small fraction of the number of incident particles rebound back.

1. The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

2. A cylinder contains hydrogen gas at pressure of 249 kPa and temperature $27^circ,C$. Its density is :$(R = 8.3,J,mol^{-1}K^{-1}$)

13.6 Ev Binding Energy

3. The solids which have negative temperature coefficient of resistance are :

4. The energy equivalent of 0.5 g of a substance is:

5. The Brewsters angle $i_b$ for an interface should be:

6. Two cylinders A and B of equal capacity are connected to each other via a stop clock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is:

7. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitch of the screw gauge is:

8. For which one of the following, Bohr model is not valid?

9. A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

10. The refractive index of a particular material is 1.67 for blue light, 1.65 for yellow light and 1.63 for red light. The dispersive power of the material is .........

In physics and chemistry, the Lyman series is a hydrogen spectral series of transitions and resulting ultravioletemission lines of the hydrogenatom as an electron goes from n ≥ 2 to n = 1 (where n is the principal quantum number), the lowest energy level of the electron. The transitions are named sequentially by Greek letters: from n = 2 to n = 1 is called Lyman-alpha, 3 to 1 is Lyman-beta, 4 to 1 is Lyman-gamma, and so on. The series is named after its discoverer, Theodore Lyman. The greater the difference in the principal quantum numbers, the higher the energy of the electromagnetic emission.

History[edit]

The first line in the spectrum of the Lyman series was discovered in 1906 by Harvard physicist Theodore Lyman, who was studying the ultraviolet spectrum of electrically excited hydrogen gas. The rest of the lines of the spectrum (all in the ultraviolet) were discovered by Lyman from 1906-1914.The spectrum of radiation emitted by hydrogen is non-continuous or discrete. Here is an illustration of the first series of hydrogen emission lines:

Historically, explaining the nature of the hydrogen spectrum was a considerable problem in physics. Nobody could predict the wavelengths of the hydrogen lines until 1885 when the Balmer formula gave an empirical formula for the visible hydrogen spectrum. Within five years Johannes Rydberg came up with an empirical formula that solved the problem, presented first in 1888 and in final form in 1890. Rydberg managed to find a formula to match the known Balmer series emission lines, and also predicted those not yet discovered. Different versions of the Rydberg formula with different simple numbers were found to generate different series of lines.

On December 1, 2011, it was announced that Voyager 1 detected the first Lyman-alpha radiation originating from the Milky Way galaxy. Lyman-alpha radiation had previously been detected from other galaxies, but due to interference from the Sun, the radiation from the Milky Way was not detectable.^[1]

The Lyman series[edit]

The version of the Rydberg formula that generated the Lyman series was:^[2]

${\displaystyle \frac{1}{\lambda }=R_{\text{H}}\left(1-\frac{1}{n^2}\right)\left(R_{\text{H}}\approx 1.0968\times {10}^7{\text{m}}^{-1}\approx \frac{13.6\text{eV}}{hc}\right)}$

Where n is a natural number greater than or equal to 2 (i.e., n = 2, 3, 4, …).

Therefore, the lines seen in the image above are the wavelengths corresponding to n = 2 on the right, to n = ∞ on the left. There are infinitely many spectral lines, but they become very dense as they approach n = ∞ (the Lyman limit), so only some of the first lines and the last one appear.

The wavelengths in the Lyman series are all ultraviolet:

Explanation and derivation[edit]

In 1914, when Niels Bohr produced his Bohr model theory, the reason why hydrogen spectral lines fit Rydberg's formula was explained. Bohr found that the electron bound to the hydrogen atom must have quantized energy levels described by the following formula,

${\displaystyle E_n=-\frac{m_e{e}^4}{2(4\pi {\epsilon }_0\hslash {)}^2}\frac{1}{n^2}=-\frac{13.6\text{eV}}{n^2}.}$

According to Bohr's third assumption, whenever an electron falls from an initial energy level E_i to a final energy level E_f, the atom must emit radiation with a wavelength of

${\displaystyle \lambda =\frac{hc}{E_{\text{i}}-E_{\text{f}}}.}$

There is also a more comfortable notation when dealing with energy in units of electronvolts and wavelengths in units of angstroms,

${\displaystyle \lambda =\frac{12398.4\text{eV}}{E_{\text{i}}-E_{\text{f}}}}$Å.

Replacing the energy in the above formula with the expression for the energy in the hydrogen atom where the initial energy corresponds to energy level n and the final energy corresponds to energy level m,

13.6 Ev Hydrogen

${\displaystyle \frac{1}{\lambda }=\frac{E_{\text{i}}-E_{\text{f}}}{12398.4\text{eV Å}}=R_{\text{H}}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)}$

Where R_H is the same Rydberg constant for hydrogen from Rydberg's long known formula. This also means that the inverse of the Rydberg constant is equal to the Lyman limit.

For the connection between Bohr, Rydberg, and Lyman, one must replace m with 1 to obtain

${\displaystyle \frac{1}{\lambda }=R_{\text{H}}\left(1-\frac{1}{n^2}\right)}$

13.6 Ev To Joules

which is Rydberg's formula for the Lyman series. Therefore, each wavelength of the emission lines corresponds to an electron dropping from a certain energy level (greater than 1) to the first energy level.

See also[edit]

13.6 Ev To Erg

References[edit]

1. ^'Voyager Probes Detect 'invisible' Milky Way Glow'. National Geographic. December 1, 2011. Retrieved 2013-03-04.CS1 maint: discouraged parameter (link)
2. ^Brehm, John; Mullin, William (1989). Introduction to the Structure of Matter. John Wiley & Sons. p. 156. ISBN0-471-60531-X.
3. ^ ^abcdefghijKramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2019). NIST Atomic Spectra Database (ver. 5.7.1), [Online]. Available: https://physics.nist.gov/asd [2020, April 11]. National Institute of Standards and Technology, Gaithersburg, MD. DOI: https://doi.org/10.18434/T4W30F