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Sergio [31]
3 years ago
14

When cs-137 decays, it emits gamma radiation. the energy of one photon is 1.06 × 10-13 j. what is the wavelength of this radiati

on?
Physics
1 answer:
Kryger [21]3 years ago
3 0
To find the solution to the problem, we would be using Planck's equation which is E = hv
Where:
E = energy
h = Planck's constant = 6.626 x 10-34 J·s
ν = frequency
Then, you’ll need a second equation which is c = λν
Where:
c = speed of light = 3 x 108 m/sec
λ = wavelength
ν = frequency
Reorder the equation to solve for frequency:ν = c/λ
Next, substitute frequency in the first equation with c/λ to get a formula you can use:
E = hν
E = hc/λ
But we are looking for the wavelength, so rearrange it more, then our final equation would be:
λ = hc / E 
λ = (6.625E-34)(3.0E8 m/s) / (1.06E-13) 
λ = 1.875E-12 m
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A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.
Mila [183]

Answer:

The smallest diameter is D =122 \ m

Explanation:

From the question we are told that

       The resolution of the telescope is \theta  =  0.00350 \ arc \ second

           The wavelength is  \lambda = 1.70 \mu m = 1.70 *10^{-6} \ m

From the question we are told that

        1 arc \ sec = \frac{1}{3600^o}

So      0.00350 \ arc \ second = x

Therefore

             x =  0.00350  *  \frac{1}{3600 }

              x = ( 9.722*10^{-7} )^o

Now  1^o  =  \frac{\pi}{180}

   So  (9.722*10^{-7})^o =  \theta

  =>    \theta  =  (9.722*10^{-7}) * \frac{\pi}{180}

           \theta  =  1.69*10^{-8} rad

The smallest diameter is mathematically represented  as

          D = \frac{1.22 \lambda }{\theta  }

substituting values

           D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8}  }

           D =122 \ m

   

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3 years ago
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