एक स्वतन्त्र हाइड्रोजन परमाणु ,`lambda_(a)`, तरंगदैर्ध्य के एक फोटॉन को अवशोषित करके n = 1 अवस्था से n = 4 अवस्था में चला जाता है । इसके तुरन्त बाद परमाणु , `lambda_(e)` , तरंगदैर्ध्य का एक फोटॉन उत्सर्जित करते हुए n = m अवस्था में आ जाता है ।
मान लीजिए कि अवशोषण तथा उत्सर्जन के दौरान परगाणु के संवेग में परिवर्तन क्रमशः `alphap_(a)` , तथा `alphap_(e)` . है । यदि `lambda_(a)/lambda_(e)=1/5` तब निम्नलिखित विकल्पों में से कौन - सा ( से ) सही है ( हैं ) ?
hc = 1242 eV nm , 1 nm = `10^(-9)' m , नियतांक और प्रकाश की गति है ।

Options

A. `lambda_c` = 418 nm
B. The ratio of kinetic energy of the electron in the state n=m to the state n=1 is `1/4`
C. m=2
D. `DeltaP_a//DeltaP_e=1/2`

Answer

B, C

Explanation

Energy of the absorbed photon in terms of its wavelength can be written as follows :
`(hc)/lambda_a=13.6[1/1^2-1/4^2]`….(i)
Energy of the emitted photon in terms of its wavelength can be written as follows :
`(hc)/lambda_e=13.6[1/m^2-1/4^2]` ....(ii)
On dividing (ii)/(i), we get the following
`lambda_a/lambda_e=([1/m^2-1/16])/([1-1/16])`
Above ratio is given to be 1/5 in this question .
`lambda_a/lambda_e=([1/m^2-1/16])/([1-1/16])=1/5`
On solving the above equation we get m=2. Hence option (c) is correct .
On substituting value of m=2 in equation (ii) and solving , we get `lambda_e` =4862Å. hence option (a) is wrong.
Further we know that , `KE_n prop z^2/n^2`
`rArr (KE_2)/(KE_1)=1/2^2=1/4` , Hence option (b) is correct .
Change in linear momentum can be written as follows for absorption and emission.
`DeltaP_a=h/lambda_a` and `DeltaP_e=h/lambda_e`
`rArr (DeltaP_a)/(DeltaP_e)=lambda_e/lambda_a=5`
Hence option (d) is wrong .