Question

In a particular metal, the K-shell has an energy of -(24.0+A) keV, while the L-shell has an energy of –(2.50+B) keV. Find the wavelength of the Ka (alpha) characteristic x-ray for this metal. Give your answer in picometers (pm)and with 3 significant figures.

A: 1

B: 4

In the Bohr model for hydrogen, the energy levels are given by E_n = (-13.6eV) / n². Find the energy of a photon emitted when an electron in a hydrogen atom jumps from n = (A+3) to n = 2.

A: 4

In an atom, an electron is jumping from a state with an energy of –(2.45+(A/100)) eV to a state with an energy of –(5.24+(B/100)) eV. Find the wavelength of the emitted photon.

A: 41

B: 6

Answer 1

Theory: (n=5) Ezan (n=4 in shell (n=3) L shell (n=2) Lp LAY & KB & Shell (121) When electron jump from L Shell emitted K Shelly photon. energy to X-rang Energy 4 Ka line Eka he 2 Constant = (E - Ex) [his plank's h = 6.626x1039 2 ho EL- Ex to J.S Giren information - Energy of L-shell E = - ( 29.0+A) Kev -- (24.0+1) ker A21 =-25. ker Energy of Shell, E = -(2.50 +B) Kev (2.50+4) ker 6.50 ker.

Answer is We from equation Know he A EL- ER -19 = 10 + 1.602410 I kev=10²er 6.626x1039 63x108 3-6-30 -(-35)]^>50278 1762 1.682x10-16 g -10 m 0.670715x10 = 0.6707AO of Ka -12 Wavelength characte ristic x = 67.07 x 10 67.07 pm 2 Pm 67. I three in 2 significant figure.

Part B of hydrogen atom (z=1) Energy for th orbit is of atom ma mass e 9 m e En = h 2 plank's 8 tonn nu Constant fo me Permitivity in free Space En - or 8E2hr to final orbit ng For initial orbit ni transition, 2 + 13.6 (1 nar (1 gr - 1 1 in E Giren information : ni (A +3) = 4+3 27 [ A: 4 2 nf = the we Answer: from equation @ get of emitted photon, energy 1 er E = 13.6 72 22 3.12 er Energy of emitted photon E=3.12 ev

Parte : an is If electron jumping from state with energy E, ev to the State with energy Ezer. this energy This ergy difference difference E,- € photon of wavelength is e mitted by A he E - Ez त्र 3 he GĘ T on, Given information or E = (2.45 + 11/100) {ts A = 417 - 2.86 en B = 26] Ez 1 (11 - (5.29+ 6/10) [As 5.30 ev Answer 3 From equation T 8 no .EE 6.62 6x1034 3x10 m S - 2.86 2.86 -5130)} X[1602 X10-19 5.0853.710-7 5085.3 AC Wavelength of emitted of emitted photon a=508503 A m