Home
Class 11
CHEMISTRY
Neon gas emits at 616 nm. The distance t...

Neon gas emits at 616 nm. The distance travelled by this radiation in 30 sec is.

A

`9 xx 10^(7)` m

B

`9 xx 10^(9)` m

C

`4.5 xx 10^(9)` m

D

`7 xx 10^(9)` m

Text Solution

Verified by Experts

The correct Answer is:
B
`v= (d)/(t ) rArr d = v xx t [ because v = 3xx10^(8) m//sec]`
Promotional Banner

Topper's Solved these Questions

  • STRUCTURE OF ATOM

    NARAYNA|Exercise EXERCISE - IV PLANCK.S QUANTUM THEORY & PHOTO ELECTRIC EFFECT|10 Videos

  • STRUCTURE OF ATOM

    NARAYNA|Exercise EXERCISE - IV H-SPECTRUM|7 Videos

  • STRUCTURE OF ATOM

    NARAYNA|Exercise EXERCISE - III|32 Videos

  • STATES OF MATTER

    NARAYNA|Exercise A & R TYPE QUESTIONS|16 Videos

  • THERMODYNAMICS

    NARAYNA|Exercise Assertion- Reason|5 Videos

Similar Questions

Explore conceptually related problems

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm , calculate a. The frequency of emission, b. The distance travelled by this radiation in 30 s c. The energy of quantum and d. The number of quanta present if it produces 2 J of energy.

The velocity versus time graph of a body moving in a straight line is as follows. The distance travelled by the body is 5 sec is

Find out distance travelled by the block in 10 sec. for a given situation

A particle is moving on X- axis. Its velocity time graph is given in the figure. Find distance travelled by the particle in 5 sec .

In the given v-t graph, the distance travelled by the body in 5 sec will be :

A particle of mass 4 kg is acted upon by steady force of 4 N.Distance travelled by the particle in 4 sec is

A body of mass 2kg at rest is supplied constant power 1J/sec, the distance travelled by body after 6 sec is

In the above question , calculate (i) distance travelled from O to A . (ii) distance travelled from B to C . (iii) total distance travelled by the body in 16 sec .