Knowee
Questions
Features
Study Tools

The energy (in eV) required to excite an electron from n=2 to n=4 state in hydrogen atom is−0.85+4.25−3.4+2.55

Question

The energy (in eV) required to excite an electron from n=2 to n=4 state in hydrogen atom is−0.85+4.25−3.4+2.55

🧐 Not the exact question you are looking for?Go ask a question

Solution

The energy required to excite an electron from n=2 to n=4 state in a hydrogen atom can be calculated by adding the given energy values.

The given energy values are -0.85 eV, 4.25 eV, -3.4 eV, and 2.55 eV.

Step 1: Add the energy values

-0.85 eV + 4.25 eV - 3.4 eV + 2.55 eV = 2.55 eV

So, the energy required to excite an electron from n=2 to n=4 state in a hydrogen atom is 2.55 eV.

Similar Questions

A hydrogen atom is in an excited state with n = 2. Using the Bohr model of hydrogen, find the following.the radius of the electron orbit (in nm).Answer for part 1the angular momentum of the electron - use SI unitsAnswer for part 2the kinetic energy of the electron (in eV).Answer for part 3the potential energy of the electron (in eV).  Note that this may be a negative number.Answer for part 4the total energy of the electron (in eV). Note that this may be a negative number.

The ionization energy of a hydrogen-like atom is 14.4 eV. The amount of energy released whenelectron jumps from the fourth orbit to the first orbit in this atom is (in eV)

A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. Thisexcited atom can make a transition to the first excited state by successively emitting two photonsof energies 10.20 and 17.00 eV, respectively. Alternatively, the atoms from the same excited statecan make a transition to the second excited state by successively emitting two photons of energies4.25 and 5.95 eV, respectively.a) The value of ‘n’ and ‘Z’ respectively are –b) How much energy will be emitted when an electron in this atom moves from (n + 1) toground state?

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:=E−Ryn2In this equation Ry stands for the Rydberg energy, and n stands for the principal quantum number of the orbital that holds the electron. (You can find the value of the Rydberg energy using the Data button on the ALEKS toolbar.)Calculate the wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron from an orbital with =n5 to an orbital with =n3. Round your answer to 3 significant digits.μm

The excitation energy of an electron from second orbit to third orbit of a hydrogen-like atomor ion with +Ze nuclear charge is 47.2eV . If the energy of H-atom in the lowest energy state is -13.6 eV , then the value of Z is

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.