Verified Solution • 2025–26 Session moderate

QAIPMT-2010-Ph-1983: NEET 2010 Physics - Class 11 AIPMT-2010-Ph-1983 Solution: The electron in the hydrogen atom jumps from excited state (n = …

The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV the stopping potential is estimated to be (the energy of the electron in nth state En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\ \text{eV}) 5.1 V 12.1 V 17.2 V 7 V

Multiple Choice Options

  • A GmM (1R12−1R22)\left( {{1 \over {R_1^2}} - {1 \over {R_2^2}}} \right) ( R 1 2 _ 1 _ _ R 2 2 _ 1 _ )
  • B GmM (1R1−1R2)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) ( R 1 _ 1 _ _ R 2 _ 1 _ )
  • C 2GmM (1R1−1R2)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) ( R 1 _ 1 _ _ R 2 _ 1 _ )
  • D 12{1 \over 2} 2 1 _ GmM (1R1−1R2)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) ( R 1 _ 1 _ _ R 2 _ 1 _ )

Step-by-Step Verified Solution

Total energy of the orbiting satellite of mass m having orbital radius r is E = _ G M m 2 r where M is the mass of the planet. Additional kinetic energy required to transfer the satellite from a circular orbit of radius R 1 to another radius R 2 is E 2 _E 1 = _ G M m 2 R E _ ( _ G M m 2 R 1 ) = _ G M m 2 R 2 + G M m 2 R 1 = G M m 2 ( 1 R 1 _ 1 R 2 )

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