Question

A maser is a laser-type device that produces electromagnetic waves with frequencies in the microwave and radio-wave bands of the electromagnetic spectrum. You can use the radio waves generated by a hydrogen maser as a standard of frequency. The frequency of these waves is 1,420,405,751.786 hertz. (A hertz is another name for one cycle per second.) A clock controlled by a hydrogen maser is off by only 1 s in 100,000 years. (The large number of significant figures given for the frequency simply illustrates the remarkable accuracy to which it has been measured.)A. What is the time for one cycle of the radio wave? Express your answer to three significant figures and include the appropriate units.B. How many cycles occur in 4 h ? Express your answer using three significant figures.C. How many cycles would have occurred during the age of the earth, which is estimated to be 4.6×109 years? Express your answer using three significant figures.D. By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth? Express your answer using three significant figures.

Answer 1

Answer:

a)  t₀ = 0.704 10⁻¹² s,  b)   #cycle = 2.05 10¹⁶ cycles ,  

c)    #_cycles = 7,418 10³² cycles,  d)   Δt = 4.6 10⁴ s

Explanation:

For this exercise we must use the relationship between the period and the frequency

       f = 1 / T

       T = 1 / f

let's apply this equation to our case

a) The time of a cycle is

       t = 1 / 1.420405 10¹²

       t₀ = 0.704 10⁻¹² s

b) number of cycles in 4 h

       t = 4 h (3600 s / 1h) = 14400 s

Let's use a direct rule of proportions (rule of three). If there is 1 cycle in t₀, how much cycle is there

      #_cycles = t (1 / t₀)

      #_cycle = 14400 s (1 cycle / 0.704 10⁻¹² s)

      #_cycle = 2.0454 10¹⁶ cycles

      #cycle = 2.05 10¹⁶ cycles

c) cycles since the age of the Earth

     t = 4.6 10⁹ years (365 days / year) (24 h / 1 day) (3600 s / 1 h)

     t = 5.2224 10²⁰ s

    #_cycles = t (1 / to)

    # _cycles = 5.2224 10²⁰ (1 / 0.704 10⁻¹²)

    #_cycles = 7,418 10³² cycles

d) Let's use the direct proportion rule. If you have a lag of Δt₁ = 1s at t₁ = 1 10⁵ years, what is the lag (Δt) at t = 4.6 10⁹ years

        Δt = t (Δt₁ / t₁)

        Δt = 4.6 10⁹ years (1 s / 10⁵ years)

        Δt = 4.6 10⁴ s