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Determine the end (final) value of n in ahydrogen atom transition if the electro

ID: 675626 • Letter: D

Question

Determine the end (final) value of n in ahydrogen atom transition if the electron starts at n = 2 and theatom absorbs a photon of light with an energy of 2.044 x 10-18J.




The speed of sound in air is 344 m/s at room temperature. Thelowest frequency of a large organ pipe is 30.0 s-1 and the highestfrequency of a piccolo is 1.50 x 104 s-1. Find the difference inwavelength between these two sounds. Determine the end (final) value of n in ahydrogen atom transition if the electron starts at n = 2 and theatom absorbs a photon of light with an energy of 2.044 x 10-18J.




The speed of sound in air is 344 m/s at room temperature. Thelowest frequency of a large organ pipe is 30.0 s-1 and the highestfrequency of a piccolo is 1.50 x 104 s-1. Find the difference inwavelength between these two sounds.

Explanation / Answer

    We Know trhat :       According to Max-planckRelationship :           E = n hv                               Given :                                           E = 2.044 x 10-18 J                                             n1 = ?                                              n2 = 2                                                h = 6.627 x 10-34 J-s                                                                                     R = 1.0973 x 1013 m-1                                                      C = 3.0 x 108 m s-1                                                                       E = hc R ( 1 / n1 2 - 1 / n22 )                Substituting the above values in the equation we get :                                             2.044x 10-18 J =   6.627 x10-34 J-s x 3.0 x 108 ms-1 x 1.0973 x 107 m-1 ( 1/  n1 2 - 1/ 22 )                                                     n1 =   1             For the given values we cannot calculate final state of n . So by consdering the given value is for n2 ifwe calculate n1 we obtain the value as 1.                  We Know that               For electromagneticRadiation :                      =    V /                   2   = 344 m / s /   30 s-1                          = 11.47 m                   1    =   344 m / s /   1.5 x 104 s-1                           =  0.0229 m     The difference in wavelengthsis                   2 - 1  = 11.4471 m.                          = 11.47 m                   1    =   344 m / s /   1.5 x 104 s-1                           =  0.0229 m     The difference in wavelengthsis                   2 - 1  = 11.4471 m.                       
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