Question

The total energy of the electron in the first excited state of hydrogen is –3.4 eV. What is the kinetic energy of the electron in this state?  

Solution

Correct option is

+3.4 eV

The kinetic and potential energies of an electron in the nth excited state are given by  

            

  

adding (i) and (ii) we get the total energy E which is

           

               

Notice from (i) and (iii) that E = –KE. Given E = –3.4 eV.

Hence    KE = –E = –(–3.4) = +3.4 eV. 

SIMILAR QUESTIONS

Q1

If the half lives of a radioactive element that decays in 6 days is 7/8. The fraction that decays in 10 days will be

Q2

The wavelength  of de Broglie waves associated with an electron (massm, charger e) accelerated through a potential difference of V is given by (his Planck’s constant)  

Q3

A proton, when accelerated through a potential difference of V volts, has a wavelength  associated with it. If an alpha particle ius to have the same wavelength , it must be accelerated through a potential difference of

Q4

A proton and an electron move with the same velocity. The associated wavelength for proton is

Q5

Two particles of masses m and 2m have equal kinetic energies. Their de Broglie wavelength are in the ratio of 

Q6

The wavelength of de Broglie wave associated with a thermal neutron of mass m at absolute temperature T is given by (here k is the Boltzmann constant) 

Q7

The momentum of a photon of frequency v is

Q8

Which energy state of the triply ionized beryllium (Be+++) has the same electrons orbital radius as that of the ground state of hydrogen? Given Zfor beryllium = 4.

Q9

Which energy state of doubly ionized lithium (Li++) has the same energy as that of the ground state of hydrogen? Given Z for lithium = 3.

Q10

The energy of the electron of hydrogen orbiting in the stationary orbit of radius rn is proportional to