Home
Class 12
CHEMISTRY
An element undergoes a reaction as show...

An element undergoes a reaction as shown `sx+2e^(-)tox^(-2)`
Energy released `=30.87` ev/atom. If the energy released is used to dissociated `4g` to `H_(2)` molecules equally into `H^(+)` and `H^(+)` is excited state of `H` atoms where the electron travels in orbit whose circumference equal to four times its de -roglie's wavelength. Determine the minimum number of moles of `x` that would be required.
Given IE of `H=13.6` ev/atom, bond energy of `H_(2)=4.526`v/molecule
(a)1
(b)2
(c)3
(d)4

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
B
`2pir=4lamda,n=4`
Total energy required `+` Total energ released `=0`
`(2xx4.526"ev"xxN_(A))+(2xx13.6N_(A))+(2xx13.6xx(1-1/16)xxN_(1))-(30.87xx x xxN_(A))=0`
`x=2`
Promotional Banner

Similar Questions

Explore conceptually related problems

An element undergoes a reaction as shown : X+e^(-) to X^(-) Energy released =30.876 eV The energy released, is used to dissociate 8 g of H_2 molecules equality into H^+ and H^* , where H^* is in an excited state, in which the electron travels a path length equal to four times its debroglie wavelength. (a)Determine the least amount (moles) of 'X' that would be required. Given : I.E. of H =13.6eV/atom Bond energy of H_2 =4.526 eV/molecule.

The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

In H-atom energy of electron is determined by :

In an excited state of hydrogen like atom an electron has total energy of -3.4 eV . If the kinetic energy of the electron is E and its de-Broglie wavelength is lambda , then

In an excited state of hydrogen like atom an electron has total energy of -3.4 eV . If the kinetic energy of the electron is E and its de-Broglie wavelength is lambda , then

The velocity of an e in excited state of H-atom is 1.093 xx 10^(6)m//s , what is the circumference of this orbit?

Assertion Second orbit circumference of hydrogen atom is two times the de-Broglie wavelength of electrons in that orbit Reason de-Broglie wavelength of electron in ground state is minimum.

If the de-Broglie wavelength of an electron revolving in 2^"nd" orbit of H-atom is x, then radius of that orbit is given by :

If an electron in a hydrogen atom is in its 2^("nd") excited state, orbiting the nucleus in a stationary orbit or radius 4.65Å , then its de - Broglie wavelength is

de-Broglie wavelength of electron in 2^("nd") excited state of hydrogen atom is: [where r_(0) is the radius of 1^("st") orbit in H-atom]