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Total number of lines emitted in infrared region when electron is de-excited from 5

^{th}excited state to ground state in hydrogen atom^{2 }orbital can exist then why not dy^{2}or dx^{2}although there r only 5 maximum possible l valuesA single electron ion has charge+Ze where Z is atomic number and e is electronic charge. it requires 16.52eV to excite the electron from the second Bohr orbit to third Bohr orbit. find

- Atomic number of element

- The energy required for transition of electron from first to third orbit

- Wavelength of photon required to remove electron from first Bohr orbit to infinity

- The Kinetic Energy of electron in first Bohr orbit

calculate the energy in calorie required toproduce, from neutral He atoms, 1 mole ofa)He= ions

b) He++ ions Using Bohr's equations.

_{2}isa.0

b.1

c.2

d.3

(b) the quantum numbers of the two levels involved in the emission of these photons,

In photoelectriceffect if the thefrequency of the light used is equal to the threshold frequency of the metalthen what happens to the electron produced because if threshold frequency is equal to frequency of light then e^{-}will have no KE so wont that mean that they wont be ejecteda.initial and final excited state

b.ionization energy

c.atomic number

a) Si ( A = 30, Z = 14)

b) P ( A = 31, Z = 15)

c) O ( A = 16, Z = 8)

d) All have same density.

1. (n-1)

2. (n+1)

3. (n-2)

4. n

EXPLAIN THIS LINE

How do they came to know that whether the alpha particles are positively chargeda. initial and final excited state

b.ionization energy

c. atomic number

a) Proton and neutron

b) Proton and deuterium

c) Deuterium and alpha particle

d) Electron and gamma rays.

(Please explain ans.)

_{2}O_{7}^{2-}+ ySO_{2}-----> Cr^{3+}+ SO_{4}^{2-}the x and y in the reaction are ?

o2directly instead of addingh2o?? ..in dat balancing eqn_{2}O_{5}OH''A GAS ABSORBS A PHOTON OF WAVELENGTH OF 350 nm AND THEN RE EMITS TWO PHOTONS. IF ONE RE-EMITTED PHOTON HAS A WAVELENGTH OF 650nm, THEN CALCULATE THE ENERGY OF OTHER RE-EMITTED PHOTON?????''

^{2+}.(a) 3.4 eV

(b) 10.2 eV

(C) 30.6 eV

(D) 6.8 eV

1. n = 4

2. l = 1

3. m = 1

4. m = 2

^{-})/r^{2}=me ve^{2}/r. in this equation what does (ze) stand for?a) n=4 to n=1

b) n=2 to n=1

c) n=4 to n=2

d) n=1 to n=5

_{l}|=1 , m_{s}= -1/2 is?^{2}2s^{2}2p^{6}3s^{2}3p^{2}. IN THIS CONFIGURATION WHY DOES NOT ONE ELECTRON FROM THE 3s^{2}ORBITAL GO TO 3p^{2}ORBITAL TO ATTAIN MAXIMUM STABILITY AS ELECTRONS IN A HALF FILLED ORBITAL ARE MORE STABLE?^{-3}m. to what shell does the electron jumpExplain question number 18