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The threshold frequency of the metal of the cathode in a photoelectric cell is $ 1 \times 10^{15}\: Hz$ . When a certain beam of light is incident on the cathode, it is found that a stopping potential '4.144 V' is required to reduce the current to zero. The frequency of the incident radiation is : $ ( h = 6.63 \times 10^{-34} J-s)$

$\begin {array} {1 1} (1)\;2.5 \times 10^{15}\: Hz & \quad (2)\;2 \times 10^{15}\: Hz \\ (3)\;4.144 \times 10^{15}\: Hz & \quad (4)\;3 \times 10^{16}\: Hz \end {array}$

 

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2 Answers

Ans : (2)
$2 \times 10^{15}\: Hz$
answered Feb 27, 2014 by thanvigandhi_1
 
Give full solution of above question how to get this answer
answered Apr 30 by priyadudhbhate2000
 

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