The threshold frequency for a metal is 7 × 10 ^14 Hz. Calculate the kinetic energy of an electron emitted when a radiation of frequency 1 × 10^15 Hz hit the metal surface ?
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Answered by
254
K.E ( kinetic energy ) = h ( f - f₀ )
h ----- > plank's constant
f₀ ------> Threshold frequency
f --------> Frequency
------------------------------------------------
-------------------------------------------------------------------------
h = 6.626 × 10⁻³⁴
f₀ = 7× 10¹⁴ Hz
f = 1 × 10¹⁵ Hz
K.E = (6.626 ×10⁻³⁴ ) ₓ [ (1 × 10¹⁵) - (7× 10¹⁴)]
= 1.987 × 10⁻¹⁹ J //
h ----- > plank's constant
f₀ ------> Threshold frequency
f --------> Frequency
------------------------------------------------
-------------------------------------------------------------------------
h = 6.626 × 10⁻³⁴
f₀ = 7× 10¹⁴ Hz
f = 1 × 10¹⁵ Hz
K.E = (6.626 ×10⁻³⁴ ) ₓ [ (1 × 10¹⁵) - (7× 10¹⁴)]
= 1.987 × 10⁻¹⁹ J //
Kimmus:
Hope this helps, if any doubts persists, kindly ask it to me. Thank you :)
It really did. Nice ans. You explained it really well. Kindly grateful to u :D
My pleasure nd Thank u @Hidya ^_^
Answered by
103
K.E. = H(F - Ft )
where,
H = Plank constant
F = Frequency
Ft = Threshold frequency
Given,
H = 6.626 * 10⁻³⁴
Ft = 7 * 10¹⁴ Hz
F = 1 * 10¹⁵ Hz = 10¹⁵ Hz
K.E. = (6.626 * 10⁻³⁴) * [(1 * 10¹⁵) - (7 * 10¹⁴)]
= 1.987 * 10⁻¹⁹ J
Kinetic Energy is 1.987 * 10⁻¹⁹ J
-Meww
where,
H = Plank constant
F = Frequency
Ft = Threshold frequency
Given,
H = 6.626 * 10⁻³⁴
Ft = 7 * 10¹⁴ Hz
F = 1 * 10¹⁵ Hz = 10¹⁵ Hz
K.E. = (6.626 * 10⁻³⁴) * [(1 * 10¹⁵) - (7 * 10¹⁴)]
= 1.987 * 10⁻¹⁹ J
Kinetic Energy is 1.987 * 10⁻¹⁹ J
-Meww
Thanks
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