2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER 2000

  • question_answer
    The energy of photon is given as: \[\Delta e\]/atom = \[3.03\times {{10}^{-19}}\,J\,ato{{m}^{-1}}\] then, the wavelength \[(\lambda )\] of the photon is:                                                                                                                                                                                  

    A)                 6.56 nm    

    B)                 65.6 nm               

    C)                 656 nm  

    D)                 0,656 nm                 (Given, h(Planck's constant) \[=6.63\,\times {{10}^{-34}}\,J-s,\] c(velocity of light) \[=3.00\,\times {{10}^{8}}\,m{{s}^{-1}}\])

    Correct Answer: C

    Solution :

                    According to formula, \[E=\frac{hc}{\lambda }\]                 \[3.03\times {{10}^{-19}}=\frac{hc}{\lambda }\]                 \[\lambda =\frac{6.63\times {{10}^{-34}}\times 3.00\times {{10}^{8}}}{3.03\times {{10}^{-19}}}\]                 \[=6.56\times {{10}^{-7}}\,m\]                 \[=6.56\times {{10}^{-7}}\times {{10}^{9}}\,\text{nm}\]                 \[=6.56\times {{10}^{2}}\,nm=656\,nm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner