`Na^+` and:Ne have the same number of electrons but the ionization potential of Na is 47.3 eV arid that of Ne only `21.6 eV. Why?

Although Cr^(3+) and Co^(2+) ions have same number of unpaired electrons but the magnetic moment of Cr^(3+) is 3.87 BM and that of Co^(2+) is 4.87 BM. Why ?

The first ionisation potential of Na is 5.1 eV the value of electron gain enthalpy of Na^+ will be

Choose the number of correct statement(s) form the following: ist ionization potential of B is higher than that of Be electron affinity of O is higher than that of S [Ar] 4s^23d^3 is the electronic configuration of Mn^(+2) Iind ionisation potential of Na>ist ionisation potential of Na 1st ionisation potential of N>IInd ionisation potential of N electronegativity of Cl>electronegativity of F C rarr C^+2 change is called IInd ionisation potential of carbon energy is required to convert He rarr He^- conversion of o rarr o^2 exothermic

Which of the following pair of elements have same number of electron in their outermost shell (1)Na ,Sr (2) Se , Te (3)Mn , Fe (4) As, Bi

The quantum numbers of six electrons are give below. Arrange them in order of increasing energies. List if any of these combination(s) has/have the same energy: (1) n=4,l=2, m_(l)=2,m_(S)=-(1)/(2) (2) n=3,l=2,m_(l)=1,m_(S)=+(1)/(2) (3) n=4,l=1, m_(l)=0,m_(S)=+(1)/(2) (4) n=3,l=2,m_(l)=-2,m_(S)=-(1)/(2) (5) n=3,l=1,m_(l)=-1,m_(S)=+(1)/(2) (6) n=4, l=1, m_(l)=-1,m_(S)=+(1)/(2) .

Among the photoelectrons, emitted due to incidence of light of frequency f on a metal surface , only the electrons having the highest kinetic energy can ionise a hydrogen atom. If this experment is performed with light of frequency 5/6f , the photoelectron having the highest kinetic energy can excite a hydrogen atom and from this excited hydrogen atom a photon of wavelength 1215 Å is emitted . Determine the (ii) value of f. Given h=6.625 xx10^(-34)*s,c=3xx10^(8)*s^(-1), 1 eV=1.6 xx10^(-19)J , ionisation energy of hydrogen =13.62 eV.

Einstein's equation for photoelectric effect is E_("max") = hf - W_(0) , where h = Planck's constant = 6.625 xx 10^(-34) J.s, f = frequency of light incident on metal surface, W_(0) = work function of metal and E_("max") = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency f is less than a minimum value f_(0) or if the wavelength lamda is greater than a maximum value lamda_(0) , the value of E_("max") would be negative, which is impossible. Thus for a particular metal surface f_(0) is the threshold frequency and lamda_(0) is the threshold wavelength for photoelectric emssion to take place. Again if the collector plate is ketp at a negative potential with respect to the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becoes zero, is known as the stopping potential, the photoelectric effect stops for a potential lower than this. [velocity of light = 3xx 10^(8) m.s^(-1) , mass of an electron m = 9.1 xx 10^(-31) kg , charge of an electron, e = 1.6 xx 10^(-19)C The threshold wavelength of photoelectric effect for a metal surface is 4600 Å . Work function of the metal (in eV) is