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Wave Optics

Question

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The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 $\times 10^{- 15}$ m is

8 $\times 10^{6} V$

80V

$9 \times 10^{5} V$

9V

Solution

8 $\times 10^{6} V$

$V = \frac{1}{4 {πε}_{o}} \frac{Q}{R} = \frac{1}{4 {πε}_{o}} \frac{(Ze)}{R} = 9 \times 10^{9} \times \frac{50 \times 1.6 \times 10^{- 19}}{9.0 \times 10^{- 15}} V = 8 \times 10^{6} Volt$

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Wave Optics

The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 $\times 10^{- 15}$ m is

8 $\times 10^{6} V$

80V

$9 \times 10^{5} V$

9V

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For Daily Free Study Material Join wiredfaculty

Wave Optics

The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0×10-15m is 8×106 V 80V 9×105 V 9V

Some More Questions From Wave Optics Chapter

In junction diode, the holes are because of

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 $\times 10^{- 15}$ m is

8 $\times 10^{6} V$

80V

$9 \times 10^{5} V$

9V