(a) Find the energy in electron volts re...

(a) Find the energy in electron volts required to strip a calcium atom `(Z=20)` of its last electron, assuming the other 19 have been removed. How does this compare with the energy required to excite the `K_(alpha)` x-ray lines of Ca, which is about 3.7 keV? (b) Why there is a difference?

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(a) 5.4 keV, (b) because other electrons change the potential

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Calculate the wavelength and wave number of the spectrum, line when an electron of H atom falls from a higher energy state n = 3 in a state n = 2 also determine the energy of a photonm to ionic this atom by removing the electron from the second orbit .Compare it with the energy of photon required to ionic the atom by removing the electron from the ground state

(a) (i) Find the wavelength of the radiation required to excite thye elctron in Li(++) from the first to the third Bohr orbit. (ii) How many spectal lines are observed in the emission spectrum of the above excited system ? (b) The energy needed to detach the electron of a hydrogen-like ion in ground state in 4 rydberg. (i) What is the wavelength of the radiation emitted when the electron jumps from the first excited state ot the ground state ? (ii) What is the radius of first orbit ? (c ) A hydrogen sample is prepared in a particular excited state A. Photons of energy 2.55 eV get absorbed into the sample to take some electrons to a further excited state B. Find the quantum number of the A and B. (d) A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 eV. (i) Identify the quantum number n of the upper and the lower energy states. Find lambda .

A single electron orbits a stationary nucleus of charge + Ze , where Z is a constant and e is the magnitude of electronic charge . It requires 47.2 eV to excite . Find a the value of Z b the energy required to excite the electron from the third to the fourth Bohr orbit. the wavelength of electromagnetic rediation required to remove the electron from the first Bohr orbit to infinity. d Find the KE,PE , and angular momentum of electron in the first Bohr orbit. the redius of the first Bohr orbit [The ionization energy of hydrogen atom = 13.6 eV Bohr radius = 5.3 xx 10^(_11) m , "velocity of light" = 3 xx 10^(-8)jm s ^(-1) , Planck's constant = 6.6 xx 10^(-34)j - s]

In 1931, Pauling defined the electronegativity of an atom as the tendency of the atom to attract electrons to itself when combined in a compound. The implication is that when a covalent bond is formed, the electrons used for bonding need not be shared equally by both atoms. If the bonding electrons spend more time around one atom, that atom will have a del- charge, and consequently the other atom will have a del+ charge. In the extreme case, where the bonding electrons are round one atom all of the time, the bond is ionic. Pauling and others have attempted to relate the electronegativity difference between two atoms to the amount of ionic character in the bond between them. Mulliken In 1934, Mulliken suggested an alternative approach to electronegativity based on the ionization energy and electron affinity of an atom. Consider two atoms A and B. If an electron is transferred from A to B, forming ions A^(+) and B^(-) , then the energy change is the ionization energy of atom A(I_(A)) minus the electron affinity of atom B(E_(B)) , that is I_(A)-E_(B) . Alternatively, if the electron was transferred the other way to give B^(+) and A^(-) ions, then the energy changed would be I_(B) - I_(A) . If A^(+) and B^(-) are actually formed, then this process require less energy , and (I_(A) - E_(B)) lt (I_(B) - E_(A)) Rerranging (I_(A)+E_(A)) lt (I_(A)-E_(B)) Now with respect to electronegativity for the same change, E.N_(A) lt E.N_(B) . Thus Mulliken suggested that electronegativity is proportional to I.E + E.A and could be regarded as the average of the ionization energy and the electron affinity of an atom. Electronegativity =((I+E))/(2) Mulliken used I and E values measured in electron volts, and the values were about 2.8 times alrger than the Pauling values. It is to be noted that I.E . and E.A values are defined for singular gaseous atoms. We have X atoms of A. if all the atoms gain one electron each the energy released is aeV . If Y atoms of A lose one electron each, then the energy absorbed is beV. Then the electronegativity of A on the Mulliken scale will be:

A metal plate is placed 5m from a monchromatic ligth soure whose power output is 10^(-3) W. Consider that a given ejected photoelectron may collect its energy from a circular area of the plate as large as ten atomic diameters (10^(-9m) in radius. The energy required to remove an electron through the metal surface is about 5.0 eV. Assuming light to be a wave, how long would it take for such a 'target' to soak up this much energy from such a light source.

One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n=1 orbit, the orbit in which it has the lowest energy. When the electon is in this lowest energy orbit, the atom is said to be in its ground electronic state. If the atom receives energy from an outside source, it is possible for the electron to move ot an orbit with a higher n value, in which case the atoms is in an excited state with a higher energy. The law of conservation of energy says that we cannot create or destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, then that same amount of energy will be liberated when the electron returns to its initial state. Lyman series is observed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electrons returns to the third, fourth and fifth orbits from higher energy orbits respectively. When electrons return form n_(2) " to " n_(1) state, the number of lines in the spectrum will equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electon comes back from energy level having energy E_(2) to energy level having energy E_(1) , then the difference may be expressed in terms of energy of photon as : E_(2)-E_(1)=DeltaE, deltaE implies (hc)/(lambda) Since, h and c are constant, deltaE corresponds to definite energy. Thus, each transition from one energy level to another will produce a radiatiob of definite wavelength. This is actually Wave number of a spectral line is given by the formula barv=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) . where R is a Rydberg's constant (R=1.1xx10^(7) m^(-1)) An electron in H-atom in M-shell on de-excitation to ground state gives maximum ........... spectrum lines.

One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n=1 orbit, the orbit in which it has the lowest energy. When the electon is in this lowest energy orbit, the atom is said to be in its ground electronic state. If the atom receives energy from an outside source, it is possible for the electron to move ot an orbit with a higher n value, in which case the atoms is in an excited state with a higher energy. The law of conservation of energy says that we cannot create or destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, then that same amount of energy will be liberated when the electron returns to its initial state. Lyman series is observed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electrons returns to the third, fourth and fifth orbits from higher energy orbits respectively. When electrons return form n_(2) " to " n_(1) state, the number of lines in the spectrum will equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electon comes back from energy level having energy E_(2) to energy level having energy E_(1) , then the difference may be expressed in terms of energy of photon as : E_(2)-E_(1)=DeltaE, deltaE implies (hc)/(lambda) Since, h and c are constant, deltaE corresponds to definite energy. Thus, each transition from one energy level to another will produce a radiatiob of definite wavelength. This is actually Wave number of a spectral line is given by the formula barv=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) . where R is a Rydberg's constant (R=1.1xx10^(7) m^(-1)) The emission spectra is observed by the consequence of transition of electrons from higher energy state to ground state of He^(+) ion. Six different photons are observed during the emission spectra, then what will be the minimum wavelength during the transition?

In 1931, Pauling defined the electronegativity of an atom as the tendency of the atom to attract electrons to itself when combined in a compound. The implication is that when a covalent bond is formed, the electrons used for bonding need not be shared equally by both atoms. If the bonding electrons spend more time around one atom, that atom will have a del- charge, and consequently the other atom will have a del+ charge. In the extreme case, where the bonding electrons are round one atom all of the time, the bond is ionic. Pauling and others have attempted to relate the electronegativity difference between two atoms to the amount of ionic character in the bond between them. Mulliken In 1934, Mulliken suggested an alternative approach to electronegativity based on the ionization energy and electron affinity of an atom. Consider two atoms A and B. If an electron is transferred from A to B, forming ions A^(+) and B^(-) , then the energy change is the ionization energy of atom A(I_(A)) minus the electron affinity of atom B(E_(B)) , that is I_(A)-E_(B) . Alternatively, if the electron was transferred the other way to give B^(+) and A^(-) ions, then the energy changed would be I_(B) - I_(A) . If A^(+) and B^(-) are actually formed, then this process require less energy , and (I_(A) - E_(B)) lt (I_(B) - E_(A)) Rerranging (I_(A)+E_(A)) lt (I_(A)-E_(B)) Now with respect to electronegativity for the same change, E.N_(A) lt E.N_(B) . Thus Mulliken suggested that electronegativity is proportional to I.E + E.A and could be regarded as the average of the ionization energy and the electron affinity of an atom. Electronegativity =((I+E))/(2) Mulliken used I and E values measured in electron volts, and the values were about 2.8 times alrger than the Pauling values. It is to be noted that I.E . and E.A values are defined for singular gaseous atoms. For a reaction A(g)+B(g) rar AB(S) the enthalpy change for the recation will be [assuming AB(s) to be an ionic compound] If I.E_(A)+E.A_(A) lt I.E_(B) +E.A_(B)

One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n=1 orbit, the orbit in which it has the lowest energy. When the electon is in this lowest energy orbit, the atom is said to be in its ground electronic state. If the atom receives energy from an outside source, it is possible for the electron to move ot an orbit with a higher n value, in which case the atoms is in an excited state with a higher energy. The law of conservation of energy says that we cannot create or destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, then that same amount of energy will be liberated when the electron returns to its initial state. Lyman series is observed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electrons returns to the third, fourth and fifth orbits from higher energy orbits respectively. When electrons return form n_(2) " to " n_(1) state, the number of lines in the spectrum will equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electon comes back from energy level having energy E_(2) to energy level having energy E_(1) , then the difference may be expressed in terms of energy of photon as : E_(2)-E_(1)=DeltaE, deltaE implies (hc)/(lambda) Since, h and c are constant, deltaE corresponds to definite energy. Thus, each transition from one energy level to another will produce a radiatiob of definite wavelength. This is actually Wave number of a spectral line is given by the formula barv=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) . where R is a Rydberg's constant (R=1.1xx10^(7) m^(-1)) What transition in the hydrogen spectrum would have the same wavelength as Balmer transitio, n=4 " to "n=2 in the He^(+) spectrum?

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