## Question

### Solution

Correct option is

900 Å to 1200 Å

Lyman series is defined by wave numbers

Hence the extreme wave numbers are

.

Therefore, the wavelength limits are

.

.

#### SIMILAR QUESTIONS

Q1

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)

Q2

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Q3

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.

Q4

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.

Q5

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?

Q6

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

Q7

The energy required to excite a hydrogen atom from n = 1 to n = 2 energy state 10.2 eV. What is the wavelength of the radiation emitter by the atom when it goes back to its ground state?

Q8

The ionisation potential of the hydrogen atom is 13.6 eV. Its energy in n = 2 energy state is

Q9

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Q10

The shortest wavelength in the :Lyman Series is 911.6 Å. Then the longest wavelength in the Lyman’s Series is